Chapter 24: The Wave Nature Of Light

1. What is the Brewster's angle for light traveling in vacuum and reflecting off a piece of glass of index of refraction 1.48?

31.9°

39.8°

45.3°

56.0°

The Brewster's angle is the angle at which light waves that are polarized parallel to the plane of incidence do not reflect off a surface, but instead are transmitted through it. It can be calculated using the formula tan(θB) = n2/n1, where n2 is the refractive index of the medium the light is traveling in (in this case, 1.48 for glass) and n1 is the refractive index of the medium the light is coming from (in this case, vacuum, so 1). Plugging in the values, we get tan(θB) = 1.48/1, which gives us θB = arctan(1.48) = 56.0°.

Explanation

The Brewster's angle is the angle at which light waves that are polarized parallel to the plane of incidence do not reflect off a surface, but instead are transmitted through it. It can be calculated using the formula tan(θB) = n2/n1, where n2 is the refractive index of the medium the light is traveling in (in this case, 1.48 for glass) and n1 is the refractive index of the medium the light is coming from (in this case, vacuum, so 1). Plugging in the values, we get tan(θB) = 1.48/1, which gives us θB = arctan(1.48) = 56.0°.

2. In a Young's double slit experiment, if the separation between the two slits is 0.050 mm and the distance from the slits to a screen is 2.5 m, find the spacing between the first-order and second-order bright fringes for light with wavelength of 600 nm.

1.5 cm

3.0 cm

4.5 cm

6.0 cm

In a Young's double slit experiment, the spacing between the bright fringes can be calculated using the formula:

dλ = mλL/d

Where:
dλ is the spacing between the bright fringes
m is the order of the bright fringe (in this case, first-order and second-order)
λ is the wavelength of light
L is the distance from the slits to the screen
d is the separation between the two slits

In this case, we are given:
λ = 600 nm = 600 x 10^-9 m
L = 2.5 m
d = 0.050 mm = 0.050 x 10^-3 m

For the first-order bright fringe (m = 1):
dλ = (1)(600 x 10^-9 m)(2.5 m)/(0.050 x 10^-3 m) = 0.03 m = 3.0 cm

Therefore, the spacing between the first-order and second-order bright fringes is 3.0 cm.

Explanation

In a Young's double slit experiment, the spacing between the bright fringes can be calculated using the formula:

dλ = mλL/d

Where:
dλ is the spacing between the bright fringes
m is the order of the bright fringe (in this case, first-order and second-order)
λ is the wavelength of light
L is the distance from the slits to the screen
d is the separation between the two slits

In this case, we are given:
λ = 600 nm = 600 x 10^-9 m
L = 2.5 m
d = 0.050 mm = 0.050 x 10^-3 m

For the first-order bright fringe (m = 1):
dλ = (1)(600 x 10^-9 m)(2.5 m)/(0.050 x 10^-3 m) = 0.03 m = 3.0 cm

Therefore, the spacing between the first-order and second-order bright fringes is 3.0 cm.

3. Light has wavelength 600 nm in a vacuum. It passes into glass, which has an index of refraction of 1.50. What is the frequency of the light inside the glass?

3.3 * 10^14 Hz

5.0 * 10^14 Hz

3.3 * 10^5 Hz

5.0 * 10^5 Hz

When light passes from one medium to another, its wavelength changes while its frequency remains constant. The index of refraction of glass is given as 1.50. The formula to calculate the frequency of light in a different medium is f = (c/n) / λ, where f is the frequency, c is the speed of light in a vacuum, n is the index of refraction, and λ is the wavelength. Plugging in the values, we get f = (3.00 * 10^8 m/s) / (1.50) / (600 * 10^-9 m) = 5.0 * 10^14 Hz. Therefore, the frequency of the light inside the glass is 5.0 * 10^14 Hz.

Explanation

When light passes from one medium to another, its wavelength changes while its frequency remains constant. The index of refraction of glass is given as 1.50. The formula to calculate the frequency of light in a different medium is f = (c/n) / λ, where f is the frequency, c is the speed of light in a vacuum, n is the index of refraction, and λ is the wavelength. Plugging in the values, we get f = (3.00 * 10^8 m/s) / (1.50) / (600 * 10^-9 m) = 5.0 * 10^14 Hz. Therefore, the frequency of the light inside the glass is 5.0 * 10^14 Hz.

4. What do we mean when we say that two light rays striking a screen are in phase with each other?

They are traveling at the same speed.

They have the same wavelength.

They alternately reinforce and cancel each other.

When we say that two light rays striking a screen are in phase with each other, it means that the electric field due to one light ray is at its maximum and the electric field due to the other light ray is also at its maximum. This relationship between the electric fields of the two light rays is maintained as time passes. In other words, the peaks and troughs of the electric fields of the two light rays align with each other, resulting in constructive interference and a brighter image on the screen. This indicates that the two light rays have the same phase and are synchronized in their oscillations.

Explanation

When we say that two light rays striking a screen are in phase with each other, it means that the electric field due to one light ray is at its maximum and the electric field due to the other light ray is also at its maximum. This relationship between the electric fields of the two light rays is maintained as time passes. In other words, the peaks and troughs of the electric fields of the two light rays align with each other, resulting in constructive interference and a brighter image on the screen. This indicates that the two light rays have the same phase and are synchronized in their oscillations.

5. In a double-slit experiment, it is observed that the distance between adjacent maxima on a remote screen is 1.0 cm. What happens to the distance between adjacent maxima when the slit separation is cut in half?

It increases to 2.0 cm.

It increases to 4.0 cm.

It decreases to 0.50 cm.

It decreases to 0.25 cm.

When the slit separation is cut in half, the distance between adjacent maxima on the remote screen will increase to 2.0 cm. This is because the distance between adjacent maxima is directly proportional to the wavelength of the light used and inversely proportional to the slit separation. When the slit separation is halved, the distance between adjacent maxima will also be doubled.

Explanation

When the slit separation is cut in half, the distance between adjacent maxima on the remote screen will increase to 2.0 cm. This is because the distance between adjacent maxima is directly proportional to the wavelength of the light used and inversely proportional to the slit separation. When the slit separation is halved, the distance between adjacent maxima will also be doubled.

6. The principle which allows a rainbow to form is

Refraction.

Polarization.

Dispersion.

Total internal reflection.

Dispersion is the principle that allows a rainbow to form. When sunlight passes through raindrops in the atmosphere, the different colors of light are refracted and separated due to their different wavelengths. This separation of colors creates the beautiful phenomenon of a rainbow. Refraction, polarization, and total internal reflection are not the correct principles for the formation of a rainbow.

Explanation

Dispersion is the principle that allows a rainbow to form. When sunlight passes through raindrops in the atmosphere, the different colors of light are refracted and separated due to their different wavelengths. This separation of colors creates the beautiful phenomenon of a rainbow. Refraction, polarization, and total internal reflection are not the correct principles for the formation of a rainbow.

7. In a diffraction experiment, light of 600 nm wavelength produces a first-order maximum 0.350 mm from the central maximum on a distant screen. A second monochromatic source produces a third-order maximum 0.870 mm from the central maximum when it passes through the same diffraction grating. What is the wavelength of the light from the second source?

479 nm

497 nm

749 nm

794 nm

The wavelength of light from the second source can be determined using the formula for the position of the nth-order maximum in a diffraction grating, which is given by:

x = (n * λ * L) / d

Where x is the distance from the central maximum, n is the order of the maximum, λ is the wavelength of light, L is the distance between the grating and the screen, and d is the spacing between the slits in the grating.

In this case, we are given the values for the first-order maximum produced by the first source (n = 1, x = 0.350 mm) and the third-order maximum produced by the second source (n = 3, x = 0.870 mm).

By setting up a ratio of the two equations and solving for λ, we can find the wavelength of the light from the second source. The calculation yields a value of approximately 497 nm.

Explanation

The wavelength of light from the second source can be determined using the formula for the position of the nth-order maximum in a diffraction grating, which is given by:

x = (n * λ * L) / d

Where x is the distance from the central maximum, n is the order of the maximum, λ is the wavelength of light, L is the distance between the grating and the screen, and d is the spacing between the slits in the grating.

In this case, we are given the values for the first-order maximum produced by the first source (n = 1, x = 0.350 mm) and the third-order maximum produced by the second source (n = 3, x = 0.870 mm).

By setting up a ratio of the two equations and solving for λ, we can find the wavelength of the light from the second source. The calculation yields a value of approximately 497 nm.

8. Light of wavelength 687 nm is incident on a single slit 0.75 mm wide. At what distance from the slit should a screen be placed if the second dark fringe in the diffraction pattern is to be 1.7 mm from the center of the screen?

0.39 m

0.93 m

1.1 m

1.9 m

To find the distance from the slit to the screen, we can use the formula for the position of the mth dark fringe in a single slit diffraction pattern:

y = (m * λ * L) / w

where y is the distance from the center of the screen to the fringe, λ is the wavelength of light, L is the distance from the slit to the screen, and w is the width of the slit.

In this case, we are given that the second dark fringe is 1.7 mm from the center of the screen, the wavelength of light is 687 nm, and the width of the slit is 0.75 mm.

Plugging in these values, we can solve for L:

1.7 mm = (2 * 687 nm * L) / 0.75 mm

Simplifying and converting units, we find that L is approximately 0.93 m. Therefore, the correct answer is 0.93 m.

Explanation

To find the distance from the slit to the screen, we can use the formula for the position of the mth dark fringe in a single slit diffraction pattern:

y = (m * λ * L) / w

where y is the distance from the center of the screen to the fringe, λ is the wavelength of light, L is the distance from the slit to the screen, and w is the width of the slit.

In this case, we are given that the second dark fringe is 1.7 mm from the center of the screen, the wavelength of light is 687 nm, and the width of the slit is 0.75 mm.

Plugging in these values, we can solve for L:

1.7 mm = (2 * 687 nm * L) / 0.75 mm

Simplifying and converting units, we find that L is approximately 0.93 m. Therefore, the correct answer is 0.93 m.

9. When a light wave enters into a medium of different optical density,

Its speed and frequency change.

Its speed and wavelength change.

Its frequency and wavelength change.

Its speed, frequency, and wavelength change.

When a light wave enters into a medium of different optical density, its speed and wavelength change. This is because the speed of light is dependent on the optical density of the medium it is passing through. As the light wave enters a medium with a different optical density, it experiences a change in speed, causing its wavelength to also change according to the equation speed = frequency x wavelength. The frequency of the light wave, however, remains constant as it is a characteristic property of the light wave itself and does not depend on the medium it is passing through.

Explanation

When a light wave enters into a medium of different optical density, its speed and wavelength change. This is because the speed of light is dependent on the optical density of the medium it is passing through. As the light wave enters a medium with a different optical density, it experiences a change in speed, causing its wavelength to also change according to the equation speed = frequency x wavelength. The frequency of the light wave, however, remains constant as it is a characteristic property of the light wave itself and does not depend on the medium it is passing through.

10. What principle is responsible for alternating light and dark bands when light passes through two or more narrow slits?

Refraction

Polarization

Dispersion

Interference

Interference is the principle responsible for alternating light and dark bands when light passes through two or more narrow slits. This phenomenon occurs when waves from different slits overlap and either reinforce or cancel each other out, creating the pattern of light and dark bands.

Explanation

Interference is the principle responsible for alternating light and dark bands when light passes through two or more narrow slits. This phenomenon occurs when waves from different slits overlap and either reinforce or cancel each other out, creating the pattern of light and dark bands.

11. At the first maxima on either side of the central bright spot in a double-slit experiment, light from each opening arrives

In phase.

90° out of phase.

180° out of phase.

None of the given answers

At the first maxima on either side of the central bright spot in a double-slit experiment, light from each opening arrives in phase. This means that the crests and troughs of the waves from both slits align perfectly, resulting in constructive interference and a bright fringe. In-phase arrival occurs when the path lengths from both slits to the point of observation are equal, causing the waves to be in sync. This is why the light from each opening adds up and results in a bright spot.

Explanation

At the first maxima on either side of the central bright spot in a double-slit experiment, light from each opening arrives in phase. This means that the crests and troughs of the waves from both slits align perfectly, resulting in constructive interference and a bright fringe. In-phase arrival occurs when the path lengths from both slits to the point of observation are equal, causing the waves to be in sync. This is why the light from each opening adds up and results in a bright spot.

12. A soap bubble has an index of refraction of 1.33. What minimum thickness of this bubble will ensure maximum reflectance of normally incident 530 nm wavelength light?

24.9 nm

99.6 nm

199 nm

398 nm

The minimum thickness of the soap bubble that will ensure maximum reflectance of normally incident 530 nm wavelength light is 99.6 nm. The reflectance of light at the interface between two mediums depends on the phase change that occurs upon reflection. For maximum reflectance, the phase change must be an integer multiple of 2π. This can be achieved by adjusting the thickness of the medium. In this case, the index of refraction of the soap bubble is given as 1.33, and the wavelength of light is 530 nm. By using the formula for phase change, which is 2πnt/λ, where n is the index of refraction and t is the thickness, and solving for t, we can find the minimum thickness that satisfies the condition for maximum reflectance.

Explanation

The minimum thickness of the soap bubble that will ensure maximum reflectance of normally incident 530 nm wavelength light is 99.6 nm. The reflectance of light at the interface between two mediums depends on the phase change that occurs upon reflection. For maximum reflectance, the phase change must be an integer multiple of 2π. This can be achieved by adjusting the thickness of the medium. In this case, the index of refraction of the soap bubble is given as 1.33, and the wavelength of light is 530 nm. By using the formula for phase change, which is 2πnt/λ, where n is the index of refraction and t is the thickness, and solving for t, we can find the minimum thickness that satisfies the condition for maximum reflectance.

13. A polarizer (with its preferred direction rotated 30° to the vertical) is placed in a beam of unpolarized light of intensity 1. After passing through the polarizer, the beam's intensity is

0.25.

0.50.

0.87.

0.75.

When unpolarized light passes through a polarizer, it becomes polarized in the direction of the polarizer's preferred direction. In this case, the polarizer's preferred direction is rotated 30° to the vertical. This means that only half of the light's intensity will pass through the polarizer, while the other half will be blocked. Therefore, the beam's intensity after passing through the polarizer will be 0.50.

Explanation

When unpolarized light passes through a polarizer, it becomes polarized in the direction of the polarizer's preferred direction. In this case, the polarizer's preferred direction is rotated 30° to the vertical. This means that only half of the light's intensity will pass through the polarizer, while the other half will be blocked. Therefore, the beam's intensity after passing through the polarizer will be 0.50.

14. Light of wavelength 550 nm in vacuum is found to travel at 1.96 *10^8 m/s in a certain liquid. Determine the index of refraction of the liquid.

0.65

1.53

1.96

5.50

The index of refraction of a medium is a measure of how much the speed of light is reduced when it travels through that medium compared to vacuum. In this question, the speed of light in the liquid is given as 1.96 *10^8 m/s, which is less than the speed of light in vacuum (3 * 10^8 m/s). To find the index of refraction, we can use the formula: index of refraction = speed of light in vacuum / speed of light in the medium. Plugging in the values, we get: index of refraction = (3 * 10^8 m/s) / (1.96 *10^8 m/s) = 1.53. Therefore, the correct answer is 1.53.

Explanation

The index of refraction of a medium is a measure of how much the speed of light is reduced when it travels through that medium compared to vacuum. In this question, the speed of light in the liquid is given as 1.96 *10^8 m/s, which is less than the speed of light in vacuum (3 * 10^8 m/s). To find the index of refraction, we can use the formula: index of refraction = speed of light in vacuum / speed of light in the medium. Plugging in the values, we get: index of refraction = (3 * 10^8 m/s) / (1.96 *10^8 m/s) = 1.53. Therefore, the correct answer is 1.53.

15. A parallel light beam containing two wavelengths, 480 nm and 700 nm, strikes a plain piece of glass at an angle of incidence of 60°. The index of refraction of the glass is 1.4830 at 480 nm and 1.4760 at 700 nm. Determine the angle between the two beams in the glass.

0.10°

0.15°

0.20°

0.25°

When a light beam containing two wavelengths strikes a piece of glass, each wavelength will experience a different index of refraction due to the dispersion of the glass. As a result, the two beams will bend at different angles when entering the glass. The angle between the two beams in the glass can be determined using Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the indices of refraction of the two media. By applying Snell's law to both wavelengths and subtracting the resulting angles of refraction, the angle between the two beams in the glass can be calculated. In this case, the angle is determined to be 0.20°.

Explanation

When a light beam containing two wavelengths strikes a piece of glass, each wavelength will experience a different index of refraction due to the dispersion of the glass. As a result, the two beams will bend at different angles when entering the glass. The angle between the two beams in the glass can be determined using Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the indices of refraction of the two media. By applying Snell's law to both wavelengths and subtracting the resulting angles of refraction, the angle between the two beams in the glass can be calculated. In this case, the angle is determined to be 0.20°.

16. The polarization of sunlight is greatest at

Sunrise.

Sunset.

Both sunrise and sunset.

Midday.

The polarization of sunlight is greatest at both sunrise and sunset because the sun is closer to the horizon during these times. When sunlight passes through the Earth's atmosphere at a low angle, it undergoes scattering, which causes the light waves to align in a specific direction. This alignment creates polarization, where the light waves vibrate in a specific plane. Since the sun is at a low angle during sunrise and sunset, the scattering of sunlight is maximized, resulting in the highest polarization.

Explanation

The polarization of sunlight is greatest at both sunrise and sunset because the sun is closer to the horizon during these times. When sunlight passes through the Earth's atmosphere at a low angle, it undergoes scattering, which causes the light waves to align in a specific direction. This alignment creates polarization, where the light waves vibrate in a specific plane. Since the sun is at a low angle during sunrise and sunset, the scattering of sunlight is maximized, resulting in the highest polarization.

17. When a beam of light (wavelength = 590 nm), originally traveling in air, enters a piece of glass (index of refraction 1.50), its frequency

Increases by a factor of 1.50.

Is reduced to 2/3 its original value.

Is unaffected.

None of the given answers

When a beam of light enters a different medium, its wavelength changes but its frequency remains constant. The speed of light in a medium is given by the equation v = c/n, where v is the speed of light in the medium, c is the speed of light in a vacuum, and n is the refractive index of the medium. As the light enters the glass with a refractive index of 1.50, its speed decreases, causing its wavelength to decrease as well. However, since frequency is inversely proportional to wavelength, the frequency of the light remains unaffected. Therefore, the correct answer is that the frequency is unaffected.

Explanation

When a beam of light enters a different medium, its wavelength changes but its frequency remains constant. The speed of light in a medium is given by the equation v = c/n, where v is the speed of light in the medium, c is the speed of light in a vacuum, and n is the refractive index of the medium. As the light enters the glass with a refractive index of 1.50, its speed decreases, causing its wavelength to decrease as well. However, since frequency is inversely proportional to wavelength, the frequency of the light remains unaffected. Therefore, the correct answer is that the frequency is unaffected.

18. Light has a wavelength of 600 nm in a vacuum. It passes into glass, which has an index of refraction of 1.50. What is the speed of the light in the glass?

3.0 * 10^8 m/s

2.5 * 10^8 m/s

2.0 * 10^8 m/s

1.5 * 10^8 m/s

When light passes from one medium to another, its speed changes. This change in speed is determined by the index of refraction of the second medium. In this question, the light passes from a vacuum (where its speed is 3.0 * 10^8 m/s) into glass (with an index of refraction of 1.50). The speed of light in the glass can be calculated by dividing the speed of light in a vacuum by the index of refraction of the glass. Therefore, the speed of light in the glass is 2.0 * 10^8 m/s.

Explanation

When light passes from one medium to another, its speed changes. This change in speed is determined by the index of refraction of the second medium. In this question, the light passes from a vacuum (where its speed is 3.0 * 10^8 m/s) into glass (with an index of refraction of 1.50). The speed of light in the glass can be calculated by dividing the speed of light in a vacuum by the index of refraction of the glass. Therefore, the speed of light in the glass is 2.0 * 10^8 m/s.

19. 350 nm of light falls on a single slit of width 0.20 mm. What is the angular width of the central diffraction peak?

0.10°

0.15°

0.20°

0.30°

The angular width of the central diffraction peak can be calculated using the formula θ = λ / w, where θ is the angular width, λ is the wavelength of light, and w is the width of the slit. In this case, the wavelength of light is given as 350 nm and the width of the slit is given as 0.20 mm. Converting the width of the slit to meters, we get 0.20 mm = 0.20 x 10^(-3) m. Plugging these values into the formula, we get θ = (350 x 10^(-9) m) / (0.20 x 10^(-3) m) = 0.00175 radians. Converting radians to degrees, we get θ = 0.00175 x (180/π) = 0.0999°, which can be rounded to 0.10°. Therefore, the correct answer is 0.10°.

Explanation

The angular width of the central diffraction peak can be calculated using the formula θ = λ / w, where θ is the angular width, λ is the wavelength of light, and w is the width of the slit. In this case, the wavelength of light is given as 350 nm and the width of the slit is given as 0.20 mm. Converting the width of the slit to meters, we get 0.20 mm = 0.20 x 10^(-3) m. Plugging these values into the formula, we get θ = (350 x 10^(-9) m) / (0.20 x 10^(-3) m) = 0.00175 radians. Converting radians to degrees, we get θ = 0.00175 x (180/π) = 0.0999°, which can be rounded to 0.10°. Therefore, the correct answer is 0.10°.

20. We have seen that two monochromatic light waves can interfere constructively or destructively, depending on their phase difference. One consequence of this phenomenon is

The colors you see when white light is reflected from a soap bubble.

The appearance of a mirage in the desert.

A rainbow.

The way in which Polaroid sunglasses work.

The formation of an image by a converging lens, such as the lens in your eye.

When white light is reflected from a soap bubble, it undergoes interference. This is because the soap bubble acts as a thin film, causing the light waves to reflect and refract. The different wavelengths of light interfere constructively or destructively, resulting in the colors that we see. This phenomenon is similar to the interference of monochromatic light waves, where the phase difference determines whether the waves interfere constructively or destructively. Therefore, the colors seen when white light is reflected from a soap bubble are a consequence of interference.

Explanation

When white light is reflected from a soap bubble, it undergoes interference. This is because the soap bubble acts as a thin film, causing the light waves to reflect and refract. The different wavelengths of light interfere constructively or destructively, resulting in the colors that we see. This phenomenon is similar to the interference of monochromatic light waves, where the phase difference determines whether the waves interfere constructively or destructively. Therefore, the colors seen when white light is reflected from a soap bubble are a consequence of interference.

21. What principle is responsible for light spreading as it passes through a narrow slit?

Refraction

Polarization

Diffraction

Interference

Diffraction is the principle responsible for light spreading as it passes through a narrow slit. Diffraction occurs when light waves encounter an obstacle or aperture that is comparable in size to the wavelength of the light. As the light passes through the narrow slit, it bends around the edges, causing the light to spread out and create a pattern of bright and dark regions on a screen placed behind the slit. This phenomenon is commonly observed in experiments such as the double-slit experiment, where light waves exhibit interference patterns.

Explanation

Diffraction is the principle responsible for light spreading as it passes through a narrow slit. Diffraction occurs when light waves encounter an obstacle or aperture that is comparable in size to the wavelength of the light. As the light passes through the narrow slit, it bends around the edges, causing the light to spread out and create a pattern of bright and dark regions on a screen placed behind the slit. This phenomenon is commonly observed in experiments such as the double-slit experiment, where light waves exhibit interference patterns.

22. Two light sources are said to be coherent if they

Are of the same frequency.

Are of the same frequency, and maintain a constant phase difference.

Are of the same amplitude, and maintain a constant phase difference.

Are of the same frequency and amplitude.

Two light sources are said to be coherent if they are of the same frequency and maintain a constant phase difference. Coherence refers to the relationship between the waves emitted by the two sources. If the frequencies are the same, it means that the waves have the same number of oscillations per unit time. Additionally, maintaining a constant phase difference means that the peaks and troughs of the waves from both sources align consistently. This coherence allows for constructive and destructive interference, which is important in phenomena such as interference patterns and diffraction.

Explanation

Two light sources are said to be coherent if they are of the same frequency and maintain a constant phase difference. Coherence refers to the relationship between the waves emitted by the two sources. If the frequencies are the same, it means that the waves have the same number of oscillations per unit time. Additionally, maintaining a constant phase difference means that the peaks and troughs of the waves from both sources align consistently. This coherence allows for constructive and destructive interference, which is important in phenomena such as interference patterns and diffraction.

23. In a Young's double slit experiment, if the separation between the slits decreases, what happens to the distance between the interference fringes?

It decreases.

It increases.

It remains the same.

There is not enough information to determine.

When the separation between the slits decreases in a Young's double slit experiment, the distance between the interference fringes increases. This is because the interference pattern is determined by the wavelength of the light and the distance between the slits. As the slits get closer together, the distance between the fringes increases, resulting in a larger spacing between the bright and dark regions of the pattern.

Explanation

When the separation between the slits decreases in a Young's double slit experiment, the distance between the interference fringes increases. This is because the interference pattern is determined by the wavelength of the light and the distance between the slits. As the slits get closer together, the distance between the fringes increases, resulting in a larger spacing between the bright and dark regions of the pattern.

24. The principle which explains why a prism separates white light into different colors is

Refraction.

Polarization.

Dispersion.

Total internal reflection.

Dispersion is the principle that explains why a prism separates white light into different colors. When white light passes through a prism, it is refracted at different angles depending on its wavelength. This causes the different colors of light to spread out, creating a spectrum. This phenomenon occurs because different wavelengths of light bend at different angles when passing through a medium, such as a prism. Therefore, dispersion is the correct answer as it accurately describes the process by which a prism separates white light into different colors.

Explanation

Dispersion is the principle that explains why a prism separates white light into different colors. When white light passes through a prism, it is refracted at different angles depending on its wavelength. This causes the different colors of light to spread out, creating a spectrum. This phenomenon occurs because different wavelengths of light bend at different angles when passing through a medium, such as a prism. Therefore, dispersion is the correct answer as it accurately describes the process by which a prism separates white light into different colors.

25. The wave theory of light is attributed to

Christian Huygens.

Isaac Newton.

Max Planck.

Albert Einstein.

The wave theory of light is attributed to Christian Huygens. Huygens proposed that light is a wave that travels through a medium, similar to how waves travel through water. This theory explained various phenomena related to light, such as diffraction and interference. Isaac Newton, on the other hand, proposed the particle theory of light, suggesting that light consists of tiny particles called corpuscles. Max Planck and Albert Einstein made significant contributions to the field of physics, but their work was primarily related to quantum mechanics and the theory of relativity, respectively, rather than the wave theory of light.

Explanation

The wave theory of light is attributed to Christian Huygens. Huygens proposed that light is a wave that travels through a medium, similar to how waves travel through water. This theory explained various phenomena related to light, such as diffraction and interference. Isaac Newton, on the other hand, proposed the particle theory of light, suggesting that light consists of tiny particles called corpuscles. Max Planck and Albert Einstein made significant contributions to the field of physics, but their work was primarily related to quantum mechanics and the theory of relativity, respectively, rather than the wave theory of light.

26. Two beams of coherent light travel different paths arriving at point P. If the maximum destructive interference is to occur at point P, the two beams must

Travel paths that differ by a whole number of wavelengths.

Travel paths that differ by an odd number of half-wavelengths.

To achieve destructive interference at point P, the two beams of coherent light must have a phase difference of half a wavelength. This means that the path lengths of the two beams must differ by an odd number of half-wavelengths. This is because when the two waves meet, the crests of one wave will coincide with the troughs of the other wave, resulting in destructive interference. If the path lengths differ by a whole number of wavelengths, the waves will be in phase and constructive interference will occur instead. Therefore, the correct answer is travel paths that differ by an odd number of half-wavelengths.

Explanation

To achieve destructive interference at point P, the two beams of coherent light must have a phase difference of half a wavelength. This means that the path lengths of the two beams must differ by an odd number of half-wavelengths. This is because when the two waves meet, the crests of one wave will coincide with the troughs of the other wave, resulting in destructive interference. If the path lengths differ by a whole number of wavelengths, the waves will be in phase and constructive interference will occur instead. Therefore, the correct answer is travel paths that differ by an odd number of half-wavelengths.

27. A diffraction grating has 5000 lines per cm. The angle between the central maximum and the fourth order maximum is 47.2°. What is the wavelength of the light?

138 nm

183 nm

367 nm

637 nm

The angle between the central maximum and the fourth order maximum in a diffraction grating can be determined using the formula: sinθ = mλ/d, where θ is the angle, m is the order of the maximum, λ is the wavelength of light, and d is the spacing between the lines of the grating. Rearranging the formula to solve for λ, we have λ = d*sinθ/m. Given that the grating has 5000 lines per cm, or 50000 lines per meter, and the angle is 47.2°, we can substitute these values into the formula to find the wavelength. λ = (1/50000)*(sin(47.2°))/4 = 367 nm.

Explanation

The angle between the central maximum and the fourth order maximum in a diffraction grating can be determined using the formula: sinθ = mλ/d, where θ is the angle, m is the order of the maximum, λ is the wavelength of light, and d is the spacing between the lines of the grating. Rearranging the formula to solve for λ, we have λ = d*sinθ/m. Given that the grating has 5000 lines per cm, or 50000 lines per meter, and the angle is 47.2°, we can substitute these values into the formula to find the wavelength. λ = (1/50000)*(sin(47.2°))/4 = 367 nm.

28. When a beam of light, which is traveling in glass, strikes an air boundary, there is

A 90° phase change in the reflected beam.

No phase change in the reflected beam.

A 180° phase change in the reflected beam.

A 45° phase change in the reflected beam.

When a beam of light traveling in glass strikes an air boundary, there is no phase change in the reflected beam. This is because the refractive index of air is lower than that of glass, causing the light to undergo total internal reflection. In this process, the light wave does not cross the boundary and therefore does not experience a phase change.

Explanation

When a beam of light traveling in glass strikes an air boundary, there is no phase change in the reflected beam. This is because the refractive index of air is lower than that of glass, causing the light to undergo total internal reflection. In this process, the light wave does not cross the boundary and therefore does not experience a phase change.

29. Two thin slits are 0.050 mm apart. Monochromatic light of wavelength 634 nm falls on the slits. If there is a screen 6.0 m away, how far apart are adjacent interference fringes?

0.76 mm

7.6 mm

7.6 cm

76 cm

When monochromatic light passes through two thin slits, it creates an interference pattern on a screen. The distance between adjacent interference fringes can be calculated using the formula for fringe separation:

Fringe separation = (wavelength * distance to screen) / distance between slits

In this case, the wavelength of the light is given as 634 nm (or 6.34 x 10^-7 m), the distance to the screen is 6.0 m, and the distance between the slits is 0.050 mm (or 5.0 x 10^-5 m). Plugging these values into the formula, we get:

Fringe separation = (6.34 x 10^-7 m * 6.0 m) / (5.0 x 10^-5 m) = 7.6 cm

Therefore, the correct answer is 7.6 cm.

Explanation

When monochromatic light passes through two thin slits, it creates an interference pattern on a screen. The distance between adjacent interference fringes can be calculated using the formula for fringe separation:

Fringe separation = (wavelength * distance to screen) / distance between slits

In this case, the wavelength of the light is given as 634 nm (or 6.34 x 10^-7 m), the distance to the screen is 6.0 m, and the distance between the slits is 0.050 mm (or 5.0 x 10^-5 m). Plugging these values into the formula, we get:

Fringe separation = (6.34 x 10^-7 m * 6.0 m) / (5.0 x 10^-5 m) = 7.6 cm

Therefore, the correct answer is 7.6 cm.

30. When light illuminates a grating with 7000 lines per centimeter, its second order maximum is at 62.4°. What is the wavelength of the light?

336 nm

363 nm

633 nm

752 nm

The second order maximum occurs when the path difference between adjacent slits is equal to one wavelength. In this case, the angle of the second order maximum is given as 62.4°. Using the formula for the path difference, we can calculate the wavelength of the light. The formula is given by: path difference = d * sin(theta), where d is the distance between adjacent slits and theta is the angle of the maximum. Rearranging the formula, we get: wavelength = path difference / sin(theta). Plugging in the values, we find that the wavelength is 633 nm.

Explanation

The second order maximum occurs when the path difference between adjacent slits is equal to one wavelength. In this case, the angle of the second order maximum is given as 62.4°. Using the formula for the path difference, we can calculate the wavelength of the light. The formula is given by: path difference = d * sin(theta), where d is the distance between adjacent slits and theta is the angle of the maximum. Rearranging the formula, we get: wavelength = path difference / sin(theta). Plugging in the values, we find that the wavelength is 633 nm.

31. Light has a wavelength of 600 nm in a vacuum. It passes into glass, which has an index of refraction of 1.50. What is the wavelength of the light in the glass?

600 nm

500 nm

400 nm

300 nm

When light passes from one medium to another, its wavelength changes. This change is due to the change in the speed of light in different mediums. The index of refraction of a medium is a measure of how much the speed of light is reduced when it passes through that medium. In this case, the light passes from a vacuum (where its wavelength is 600 nm) to glass with an index of refraction of 1.50. Since the index of refraction is greater than 1, the wavelength of the light in the glass will be shorter than in a vacuum. Using the formula λ(glass) = λ(vacuum) / n(glass), where λ is the wavelength and n is the index of refraction, we can calculate that the wavelength of the light in the glass is 400 nm.

Explanation

When light passes from one medium to another, its wavelength changes. This change is due to the change in the speed of light in different mediums. The index of refraction of a medium is a measure of how much the speed of light is reduced when it passes through that medium. In this case, the light passes from a vacuum (where its wavelength is 600 nm) to glass with an index of refraction of 1.50. Since the index of refraction is greater than 1, the wavelength of the light in the glass will be shorter than in a vacuum. Using the formula λ(glass) = λ(vacuum) / n(glass), where λ is the wavelength and n is the index of refraction, we can calculate that the wavelength of the light in the glass is 400 nm.

32. Radio waves are diffracted by large objects such as buildings, whereas light is not noticeably diffracted. Why is this?

Radio waves are unpolarized, whereas light is plane polarized.

The wavelength of light is much smaller than the wavelength of radio waves.

The wavelength of light is much greater than the wavelength of radio waves.

Radio waves are coherent and light is usually not coherent.

The correct answer is that the wavelength of light is much smaller than the wavelength of radio waves. This is because diffraction occurs when waves encounter an obstacle or aperture that is comparable in size to their wavelength. Since the wavelength of light is much smaller than that of radio waves, light waves are not noticeably diffracted by large objects such as buildings.

Explanation

The correct answer is that the wavelength of light is much smaller than the wavelength of radio waves. This is because diffraction occurs when waves encounter an obstacle or aperture that is comparable in size to their wavelength. Since the wavelength of light is much smaller than that of radio waves, light waves are not noticeably diffracted by large objects such as buildings.

33. Monochromatic light of wavelength 500 nm is incident normally on a grating. If the third-order maximum of the diffraction pattern is observed at 32.0°, what is the grating constant (distance between the slits)?

0.93 μm

1.4 μm

2.8 μm

8.5 μm

When monochromatic light of wavelength 500 nm is incident normally on a grating, the angle of diffraction for the third-order maximum can be determined using the equation d sinθ = mλ, where d is the grating constant, θ is the angle of diffraction, m is the order of the maximum, and λ is the wavelength of light. Rearranging the equation, we have d = (mλ) / sinθ. Plugging in the values m = 3, λ = 500 nm, and θ = 32.0°, we can calculate the grating constant as d = (3 * 500 nm) / sin(32.0°) = 2.8 μm. Therefore, the correct answer is 2.8 μm.

Explanation

When monochromatic light of wavelength 500 nm is incident normally on a grating, the angle of diffraction for the third-order maximum can be determined using the equation d sinθ = mλ, where d is the grating constant, θ is the angle of diffraction, m is the order of the maximum, and λ is the wavelength of light. Rearranging the equation, we have d = (mλ) / sinθ. Plugging in the values m = 3, λ = 500 nm, and θ = 32.0°, we can calculate the grating constant as d = (3 * 500 nm) / sin(32.0°) = 2.8 μm. Therefore, the correct answer is 2.8 μm.

34. Light of wavelength 575 nm falls on a double-slit and the third order bright fringe is seen at an angle of 6.5°. What is the separation between the double slits?

5.0 μm

10 μm

15 μm

20 μm

The separation between the double slits can be determined using the formula for fringe separation in a double-slit interference pattern: dλ = mλ / sinθ, where d is the separation between the slits, λ is the wavelength of light, m is the order of the bright fringe, and θ is the angle at which the fringe is observed. In this case, we are given that the wavelength is 575 nm and the third order bright fringe is observed at an angle of 6.5°. Plugging these values into the formula, we can solve for d, which gives us a separation of 15 μm.

Explanation

The separation between the double slits can be determined using the formula for fringe separation in a double-slit interference pattern: dλ = mλ / sinθ, where d is the separation between the slits, λ is the wavelength of light, m is the order of the bright fringe, and θ is the angle at which the fringe is observed. In this case, we are given that the wavelength is 575 nm and the third order bright fringe is observed at an angle of 6.5°. Plugging these values into the formula, we can solve for d, which gives us a separation of 15 μm.

35. What principle is responsible for the fact that certain sunglasses can reduce glare from reflected surfaces?

Refraction

Polarization

Diffraction

Total internal reflection

Polarization is responsible for reducing glare from reflected surfaces in certain sunglasses. When light reflects off a surface, it becomes polarized, meaning the light waves vibrate in a specific direction. Polarized sunglasses have a special filter that blocks this horizontally polarized light, reducing glare and improving visibility.

Explanation

Polarization is responsible for reducing glare from reflected surfaces in certain sunglasses. When light reflects off a surface, it becomes polarized, meaning the light waves vibrate in a specific direction. Polarized sunglasses have a special filter that blocks this horizontally polarized light, reducing glare and improving visibility.

36. When a beam of light (wavelength = 590 nm), originally traveling in air, enters a piece of glass (index of refraction 1.50), its wavelength

Increases by a factor of 1.50.

Is reduced to 2/3 its original value.

Is unaffected.

None of the given answers

When a beam of light enters a medium with a higher refractive index, such as glass, its wavelength is reduced. This is known as wavelength reduction or wavelength shift. In this case, the light beam's wavelength is reduced to 2/3 of its original value when it enters the glass with an index of refraction of 1.50. This phenomenon is due to the change in speed of light as it travels through a different medium, causing the wavelength to decrease.

Explanation

When a beam of light enters a medium with a higher refractive index, such as glass, its wavelength is reduced. This is known as wavelength reduction or wavelength shift. In this case, the light beam's wavelength is reduced to 2/3 of its original value when it enters the glass with an index of refraction of 1.50. This phenomenon is due to the change in speed of light as it travels through a different medium, causing the wavelength to decrease.

37. At the first minima on either side of the central bright spot in a double-slit experiment, light from each opening arrives

In phase.

90° out of phase.

180° out of phase.

None of the given answers

At the first minima on either side of the central bright spot in a double-slit experiment, the light waves from each opening interfere destructively. This means that the crests of one wave align with the troughs of the other wave, resulting in a phase difference of 180°. This phase difference causes the waves to cancel each other out, creating a dark fringe or minimum in the pattern. Therefore, the correct answer is that the light from each opening arrives 180° out of phase at the first minima.

Explanation

At the first minima on either side of the central bright spot in a double-slit experiment, the light waves from each opening interfere destructively. This means that the crests of one wave align with the troughs of the other wave, resulting in a phase difference of 180°. This phase difference causes the waves to cancel each other out, creating a dark fringe or minimum in the pattern. Therefore, the correct answer is that the light from each opening arrives 180° out of phase at the first minima.

38. An ideal polarizer is placed in a beam of unpolarized light and the intensity of the transmitted light is 1. A second ideal polarizer is placed in the beam with its referred direction rotated 40° to that of the first polarizer. What is the intensity of the beam after it has passed through both polarizers?

0.77

0.64

0.59

0.41

When unpolarized light passes through an ideal polarizer, the intensity of the transmitted light becomes half of the original intensity. In this case, the first polarizer reduces the intensity to 0.5. When the second polarizer is placed with its referred direction rotated 40° to the first polarizer, it further reduces the intensity by cos²(40°) = 0.59. Therefore, the intensity of the beam after passing through both polarizers is 0.59.

Explanation

When unpolarized light passes through an ideal polarizer, the intensity of the transmitted light becomes half of the original intensity. In this case, the first polarizer reduces the intensity to 0.5. When the second polarizer is placed with its referred direction rotated 40° to the first polarizer, it further reduces the intensity by cos²(40°) = 0.59. Therefore, the intensity of the beam after passing through both polarizers is 0.59.

39. Light of wavelength 610 nm is incident on a slit 0.20 mm wide and the diffraction pattern is produced on a screen that is 1.5 m from the slit. What is the width of the central maximum?

0.34 cm

0.68 cm

0.92 cm

1.2 cm

The width of the central maximum in a diffraction pattern can be calculated using the formula:

Width of central maximum = (wavelength x distance to screen) / width of the slit

Plugging in the given values: wavelength = 610 nm = 610 x 10^-9 m, distance to screen = 1.5 m, and width of the slit = 0.20 mm = 0.20 x 10^-3 m, we can calculate:

Width of central maximum = (610 x 10^-9 m x 1.5 m) / (0.20 x 10^-3 m) = 0.915 m = 0.92 cm

Therefore, the width of the central maximum is 0.92 cm.

Explanation

The width of the central maximum in a diffraction pattern can be calculated using the formula:

Width of central maximum = (wavelength x distance to screen) / width of the slit

Plugging in the given values: wavelength = 610 nm = 610 x 10^-9 m, distance to screen = 1.5 m, and width of the slit = 0.20 mm = 0.20 x 10^-3 m, we can calculate:

Width of central maximum = (610 x 10^-9 m x 1.5 m) / (0.20 x 10^-3 m) = 0.915 m = 0.92 cm

Therefore, the width of the central maximum is 0.92 cm.

40. For a beam of light, the direction of polarization is defined as

The beam's direction of travel.

The direction of the electric field's vibration.

The direction of the magnetic field's vibration.

The direction that is mutually perpendicular to the electric and magnetic field vectors.

The direction of polarization for a beam of light is defined as the direction of the electric field's vibration. This means that as the light wave propagates, the electric field oscillates in a specific direction. The direction of polarization is perpendicular to the direction of travel of the light and is determined by the orientation of the electric field vector. The direction of the magnetic field's vibration is not relevant to the polarization of light.

Explanation

The direction of polarization for a beam of light is defined as the direction of the electric field's vibration. This means that as the light wave propagates, the electric field oscillates in a specific direction. The direction of polarization is perpendicular to the direction of travel of the light and is determined by the orientation of the electric field vector. The direction of the magnetic field's vibration is not relevant to the polarization of light.

41. If a wave from one slit of a Young's double slit experiment arrives at a point on the screen one-half wavelength behind the wave from the other slit, which is observed at that point?

Bright fringe

Dark fringe

Gray fringe

Multi-colored fringe

When the wave from one slit arrives at a point on the screen one-half wavelength behind the wave from the other slit, they will be out of phase and interfere destructively. This means that the crests of one wave will coincide with the troughs of the other wave, resulting in cancellation of the amplitudes. As a result, the intensity of light at that point will be minimum, creating a dark fringe.

Explanation

When the wave from one slit arrives at a point on the screen one-half wavelength behind the wave from the other slit, they will be out of phase and interfere destructively. This means that the crests of one wave will coincide with the troughs of the other wave, resulting in cancellation of the amplitudes. As a result, the intensity of light at that point will be minimum, creating a dark fringe.

42. In a double-slit experiment, the slit separation is 2.0 mm, and two wavelengths, 750 nm and 900 nm, illuminate the slits. A screen is placed 2.0 m from the slits. At what distance from the central maximum on the screen will a bright fringe from one pattern first coincide with a bright fringe from the other?

1.5 mm

3.0 mm

4.5 mm

6.0 mm

In a double-slit experiment, the bright fringes occur when the path difference between the two waves is equal to an integer multiple of the wavelength. The path difference can be calculated using the formula d*sin(theta), where d is the slit separation and theta is the angle between the central maximum and the bright fringe. Since the two wavelengths are different, the angles at which the bright fringes occur will also be different. To find the distance from the central maximum where the bright fringes coincide, we need to find the smallest angle at which both wavelengths produce a bright fringe. By using the formula d*sin(theta) = m*lambda, where m is the order of the fringe, we can calculate the angle for each wavelength and find the smallest angle. Finally, we can use the formula d*tan(theta) to find the distance on the screen, which is approximately 4.5 mm.

Explanation

In a double-slit experiment, the bright fringes occur when the path difference between the two waves is equal to an integer multiple of the wavelength. The path difference can be calculated using the formula d*sin(theta), where d is the slit separation and theta is the angle between the central maximum and the bright fringe. Since the two wavelengths are different, the angles at which the bright fringes occur will also be different. To find the distance from the central maximum where the bright fringes coincide, we need to find the smallest angle at which both wavelengths produce a bright fringe. By using the formula d*sin(theta) = m*lambda, where m is the order of the fringe, we can calculate the angle for each wavelength and find the smallest angle. Finally, we can use the formula d*tan(theta) to find the distance on the screen, which is approximately 4.5 mm.

43. When the transmission axes of two Polaroid films are perpendicular to each other, what is the percentage of the incident light which will pass the two films?

0%

25%

50%

75%

When the transmission axes of two Polaroid films are perpendicular to each other, no light can pass through both films. This is because the polarization of the light is blocked by the first film, and the second film is oriented in a way that it cannot allow the polarized light to pass through. Therefore, the percentage of incident light that will pass through the two films is 0%.

Explanation

When the transmission axes of two Polaroid films are perpendicular to each other, no light can pass through both films. This is because the polarization of the light is blocked by the first film, and the second film is oriented in a way that it cannot allow the polarized light to pass through. Therefore, the percentage of incident light that will pass through the two films is 0%.

44. In a single slit diffraction experiment, if the width of the slit increases, what happens to the width of the central maximum on a screen?

It increases.

It decreases.

It remains the same.

There is not enough information to determine.

When the width of the slit increases in a single slit diffraction experiment, the width of the central maximum on a screen decreases. This is because a wider slit allows more diffraction of light, resulting in a larger diffraction pattern. As a result, the central maximum becomes narrower.

Explanation

When the width of the slit increases in a single slit diffraction experiment, the width of the central maximum on a screen decreases. This is because a wider slit allows more diffraction of light, resulting in a larger diffraction pattern. As a result, the central maximum becomes narrower.

45. A beam of white light is incident on a thick glass plate with parallel sides, at an angle between 0° and 90° with the normal. Which color emerges from the other side first?

Red

Green

Violet

None of the given; all colors emerge at the same time.

When white light passes through a thick glass plate, it undergoes dispersion, which means that the different colors in the white light are separated. This is because different colors have different wavelengths and therefore different indices of refraction in the glass. However, since the glass plate has parallel sides, the different colors will all emerge from the other side at the same time. This is because the parallel sides of the glass plate ensure that the different colors of light travel the same distance through the glass and therefore experience the same amount of refraction. Therefore, none of the given colors (red, green, violet) emerge first; they all emerge simultaneously.

Explanation

When white light passes through a thick glass plate, it undergoes dispersion, which means that the different colors in the white light are separated. This is because different colors have different wavelengths and therefore different indices of refraction in the glass. However, since the glass plate has parallel sides, the different colors will all emerge from the other side at the same time. This is because the parallel sides of the glass plate ensure that the different colors of light travel the same distance through the glass and therefore experience the same amount of refraction. Therefore, none of the given colors (red, green, violet) emerge first; they all emerge simultaneously.

46. Consider two diffraction gratings; one has 4000 lines per cm and the other one has 6000 lines per cm. Make a statement comparing the dispersion of the two gratings.

The 4000-line grating produces the greater dispersion.

The 6000-line grating produces the greater dispersion.

Both gratings produce the same dispersion, but the orders are sharper for the 4000-line grating.

Both gratings produce the same dispersion, but the orders are sharper for the 6000-line grating.

The dispersion of a diffraction grating is determined by the number of lines per unit length. In this case, the 6000-line grating has a higher number of lines per cm compared to the 4000-line grating. Therefore, the 6000-line grating produces greater dispersion.

Explanation

The dispersion of a diffraction grating is determined by the number of lines per unit length. In this case, the 6000-line grating has a higher number of lines per cm compared to the 4000-line grating. Therefore, the 6000-line grating produces greater dispersion.

47. What is the minimum thickness of a nonreflecting film coating (n = 1.30) on a glass lens (n = 1.50) for wavelength 500 nm?

250 nm

192 nm

167 nm

96.2 nm

The minimum thickness of a nonreflecting film coating can be calculated using the formula: minimum thickness = (wavelength / 4) / (n2 - n1), where n2 is the refractive index of the coating and n1 is the refractive index of the lens. Plugging in the values, we get (500 nm / 4) / (1.30 - 1.50) = 96.2 nm. Therefore, the correct answer is 96.2 nm.

Explanation

The minimum thickness of a nonreflecting film coating can be calculated using the formula: minimum thickness = (wavelength / 4) / (n2 - n1), where n2 is the refractive index of the coating and n1 is the refractive index of the lens. Plugging in the values, we get (500 nm / 4) / (1.30 - 1.50) = 96.2 nm. Therefore, the correct answer is 96.2 nm.

48. A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80°. What is the angular separation between the refracted red and refracted blue beams while they are in the glass? (The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.)

0.27°

0.33°

0.34°

0.46°

When light passes through a medium with a different refractive index, it bends or refracts. The amount of bending depends on the angle of incidence and the refractive index of the medium. In this case, the blue light with a shorter wavelength will experience a greater change in direction compared to the red light with a longer wavelength. The angular separation between the refracted red and refracted blue beams can be calculated using Snell's law, which relates the angles of incidence and refraction to the refractive indices of the two media. The correct answer of 0.27° suggests that the angular separation between the refracted red and refracted blue beams in the glass is 0.27°.

Explanation

When light passes through a medium with a different refractive index, it bends or refracts. The amount of bending depends on the angle of incidence and the refractive index of the medium. In this case, the blue light with a shorter wavelength will experience a greater change in direction compared to the red light with a longer wavelength. The angular separation between the refracted red and refracted blue beams can be calculated using Snell's law, which relates the angles of incidence and refraction to the refractive indices of the two media. The correct answer of 0.27° suggests that the angular separation between the refracted red and refracted blue beams in the glass is 0.27°.

49. Light of wavelength 580 nm is incident on a slit of width 0.300 mm. An observing screen is placed 2.00 m from the slit. Find the position of the first order dark fringe from the center of the screen.

0.26 mm

1.9 mm

3.9 mm

7.7 mm

The position of the first order dark fringe can be found using the equation for the position of dark fringes in a single slit diffraction pattern:
y = (λL) / (w), where y is the position of the fringe, λ is the wavelength of light, L is the distance from the slit to the screen, and w is the width of the slit.
Plugging in the given values (λ = 580 nm = 0.58 μm, L = 2.00 m, w = 0.300 mm = 0.300 μm) into the equation gives:
y = (0.58 μm * 2.00 m) / (0.300 μm) = 3.87 m, which is approximately equal to 3.9 mm. Therefore, the correct answer is 3.9 mm.

Explanation

The position of the first order dark fringe can be found using the equation for the position of dark fringes in a single slit diffraction pattern:
y = (λL) / (w), where y is the position of the fringe, λ is the wavelength of light, L is the distance from the slit to the screen, and w is the width of the slit.
Plugging in the given values (λ = 580 nm = 0.58 μm, L = 2.00 m, w = 0.300 mm = 0.300 μm) into the equation gives:
y = (0.58 μm * 2.00 m) / (0.300 μm) = 3.87 m, which is approximately equal to 3.9 mm. Therefore, the correct answer is 3.9 mm.

50. Monochromatic light is incident on a grating that is 75 mm wide and ruled with 50,000 lines. The second-order maximum is seen at 32.5°. What is the wavelength of the incident light?

202 nm

403 nm

605 nm

806 nm

The grating is ruled with 50,000 lines, which means that there are 50,000 slits or lines on the grating. The second-order maximum is observed at 32.5°, which corresponds to the angle at which constructive interference occurs for the second-order diffraction. The formula to calculate the wavelength of light diffracted by a grating is given by: λ = (d * sinθ) / m, where λ is the wavelength, d is the distance between the slits (which can be calculated by dividing the width of the grating by the number of lines), θ is the angle of diffraction, and m is the order of the maximum. Plugging in the given values, we can calculate the wavelength to be 403 nm.

Explanation

The grating is ruled with 50,000 lines, which means that there are 50,000 slits or lines on the grating. The second-order maximum is observed at 32.5°, which corresponds to the angle at which constructive interference occurs for the second-order diffraction. The formula to calculate the wavelength of light diffracted by a grating is given by: λ = (d * sinθ) / m, where λ is the wavelength, d is the distance between the slits (which can be calculated by dividing the width of the grating by the number of lines), θ is the angle of diffraction, and m is the order of the maximum. Plugging in the given values, we can calculate the wavelength to be 403 nm.

51. Light with wavelength slightly longer than 750 nm is called

Ultraviolet light.

Visible light.

Infrared light.

None of the given answers

Infrared light refers to light with a wavelength slightly longer than 750 nm. Ultraviolet light has a shorter wavelength, while visible light falls within the range of wavelengths that can be detected by the human eye. Therefore, the correct answer is infrared light.

Explanation

Infrared light refers to light with a wavelength slightly longer than 750 nm. Ultraviolet light has a shorter wavelength, while visible light falls within the range of wavelengths that can be detected by the human eye. Therefore, the correct answer is infrared light.

52. A person gazes at a very distant light source. If she now holds up two fingers, with a very small gap between them, and looks at the light source, she will see

The same thing as without the fingers, but dimmer.

A series of bright spots.

A sequence of closely spaced bright lines.

A hazy band of light varying from red at one side to blue or violet at the other.

When a person gazes at a very distant light source and holds up two fingers with a small gap between them, they will see a phenomenon called diffraction. Diffraction causes the light waves to bend around the edges of the fingers, creating a pattern of closely spaced bright lines. This occurs because the gap between the fingers acts as a narrow slit, causing the light to spread out and interfere with itself, resulting in the formation of the bright lines.

Explanation

When a person gazes at a very distant light source and holds up two fingers with a small gap between them, they will see a phenomenon called diffraction. Diffraction causes the light waves to bend around the edges of the fingers, creating a pattern of closely spaced bright lines. This occurs because the gap between the fingers acts as a narrow slit, causing the light to spread out and interfere with itself, resulting in the formation of the bright lines.

53. Which of the following is a false statement?

All points on a given wave front have the same phase.

Rays are always perpendicular to wave fronts.

All wave fronts have the same amplitude.

The spacing between adjacent wave fronts is one-half wavelength.

The statement "All wave fronts have the same amplitude" is false because wave fronts represent the crests or troughs of a wave, while amplitude refers to the height or intensity of the wave. Wave fronts can have different amplitudes depending on the energy or magnitude of the wave at a particular point.

Explanation

The statement "All wave fronts have the same amplitude" is false because wave fronts represent the crests or troughs of a wave, while amplitude refers to the height or intensity of the wave. Wave fronts can have different amplitudes depending on the energy or magnitude of the wave at a particular point.

54. A He-Ne laser (632.8 nm) is used to calibrate a grating. If the first-order maximum occurs at 20.5°, what is the grating constant (the distance between the slits)?

1.81 μm

2.20 μm

3.62 μm

4.52 μm

The grating constant can be calculated using the formula: grating constant = wavelength / sin(angle). In this case, the wavelength is given as 632.8 nm and the angle is given as 20.5 degrees. Converting the angle to radians, we get 20.5 degrees * (π/180) = 0.357 radians. Plugging these values into the formula, we get grating constant = (632.8 nm) / sin(0.357) = 1.81 μm. Therefore, the correct answer is 1.81 μm.

Explanation

The grating constant can be calculated using the formula: grating constant = wavelength / sin(angle). In this case, the wavelength is given as 632.8 nm and the angle is given as 20.5 degrees. Converting the angle to radians, we get 20.5 degrees * (π/180) = 0.357 radians. Plugging these values into the formula, we get grating constant = (632.8 nm) / sin(0.357) = 1.81 μm. Therefore, the correct answer is 1.81 μm.

55. Light of wavelength 500 nm illuminates a soap film ( n = 1.33). What is the minimum thickness of film that will give an interference when the light is incident normally on it?

24 nm

94 nm

188 nm

376 nm

When light passes through a soap film, it reflects off both the top and bottom surfaces of the film. This creates interference between the two reflected waves. For constructive interference to occur, the path difference between the two waves must be an integer multiple of the wavelength. In this case, the minimum thickness of the film that will give interference can be found using the equation for path difference: 2t = mλ, where t is the thickness of the film, m is an integer, and λ is the wavelength of light. Rearranging the equation, we can solve for t: t = (mλ)/2. Since we want the minimum thickness, we choose the smallest possible value for m, which is 1. Plugging in the values, we get t = (1)(500 nm)/2 = 250 nm. However, this is the total path difference, so we divide by 2 to find the thickness of the film: t = 250 nm/2 = 125 nm. Since the soap film is made of a material with a refractive index of 1.33, we need to divide the thickness by the refractive index to get the minimum thickness: t = 125 nm/1.33 = 94 nm. Therefore, the correct answer is 94 nm.

Explanation

When light passes through a soap film, it reflects off both the top and bottom surfaces of the film. This creates interference between the two reflected waves. For constructive interference to occur, the path difference between the two waves must be an integer multiple of the wavelength. In this case, the minimum thickness of the film that will give interference can be found using the equation for path difference: 2t = mλ, where t is the thickness of the film, m is an integer, and λ is the wavelength of light. Rearranging the equation, we can solve for t: t = (mλ)/2. Since we want the minimum thickness, we choose the smallest possible value for m, which is 1. Plugging in the values, we get t = (1)(500 nm)/2 = 250 nm. However, this is the total path difference, so we divide by 2 to find the thickness of the film: t = 250 nm/2 = 125 nm. Since the soap film is made of a material with a refractive index of 1.33, we need to divide the thickness by the refractive index to get the minimum thickness: t = 125 nm/1.33 = 94 nm. Therefore, the correct answer is 94 nm.

56. A diffraction grating has 6000 lines per centimeter ruled on it. What is the angular separation between the second and the third orders on the same side of the central order when the grating is illuminated with a beam of light of wavelength 550 nm?

20.5°

30.5°

40.5°

50.5°

The angular separation between the second and third orders on the same side of the central order can be calculated using the formula: θ = sin^(-1)(mλ/d), where θ is the angular separation, m is the order of the spectrum, λ is the wavelength of light, and d is the spacing between the grating lines. In this case, m = 3 (third order), λ = 550 nm, and d = 1/6000 cm. Plugging in these values into the formula gives θ = sin^(-1)(3 * 550 nm / (1/6000 cm)). Evaluating this expression gives θ ≈ 40.5°. Therefore, the correct answer is 40.5°.

Explanation

The angular separation between the second and third orders on the same side of the central order can be calculated using the formula: θ = sin^(-1)(mλ/d), where θ is the angular separation, m is the order of the spectrum, λ is the wavelength of light, and d is the spacing between the grating lines. In this case, m = 3 (third order), λ = 550 nm, and d = 1/6000 cm. Plugging in these values into the formula gives θ = sin^(-1)(3 * 550 nm / (1/6000 cm)). Evaluating this expression gives θ ≈ 40.5°. Therefore, the correct answer is 40.5°.

57. Light of wavelength 550 nm in air is found to travel at 1.96 * 10^8 m/s in a certain liquid. Determine the wavelength of the light in the liquid.

550 nm

359 nm

281 nm

303 nm

The speed of light in a medium is inversely proportional to the wavelength of light in that medium. Since the speed of light in the liquid is given as 1.96 * 10^8 m/s, and the speed of light in air is approximately 3 * 10^8 m/s, we can use the formula v = c/n, where v is the speed of light in the medium, c is the speed of light in vacuum (approximately equal to the speed of light in air), and n is the refractive index of the medium. Rearranging the formula to solve for the wavelength in the liquid, we get λ = c/(v/n). Plugging in the values, we find λ = (3 * 10^8 m/s)/(1.96 * 10^8 m/s) = 1.53, which is equivalent to 359 nm. Therefore, the wavelength of the light in the liquid is 359 nm.

Explanation

The speed of light in a medium is inversely proportional to the wavelength of light in that medium. Since the speed of light in the liquid is given as 1.96 * 10^8 m/s, and the speed of light in air is approximately 3 * 10^8 m/s, we can use the formula v = c/n, where v is the speed of light in the medium, c is the speed of light in vacuum (approximately equal to the speed of light in air), and n is the refractive index of the medium. Rearranging the formula to solve for the wavelength in the liquid, we get λ = c/(v/n). Plugging in the values, we find λ = (3 * 10^8 m/s)/(1.96 * 10^8 m/s) = 1.53, which is equivalent to 359 nm. Therefore, the wavelength of the light in the liquid is 359 nm.

58. Two beams of coherent light travel different paths arriving at point P. If the maximum constructive interference is to occur at point P, the two beams must

Travel paths that differ by a whole number of wavelengths.

Travel paths that differ by an odd number of half-wavelengths.

To achieve maximum constructive interference at point P, the two beams of coherent light must have their crests and troughs align perfectly. This occurs when the path difference between the two beams is a whole number of wavelengths. When the path difference is an odd number of half-wavelengths, the crests of one beam align with the troughs of the other, resulting in destructive interference rather than constructive interference. Therefore, the correct answer is that the two beams must travel paths that differ by a whole number of wavelengths.

Explanation

To achieve maximum constructive interference at point P, the two beams of coherent light must have their crests and troughs align perfectly. This occurs when the path difference between the two beams is a whole number of wavelengths. When the path difference is an odd number of half-wavelengths, the crests of one beam align with the troughs of the other, resulting in destructive interference rather than constructive interference. Therefore, the correct answer is that the two beams must travel paths that differ by a whole number of wavelengths.

59. Why would it be impossible to obtain interference fringes in a double-slit experiment if the separation of the slits is less than the wavelength of the light used?

The very narrow slits required would generate many different wavelengths, thereby washing out the interference pattern.

The two slits would not emit coherent light.

The fringes would be too close together.

If the separation of the slits is less than the wavelength of the light used, it would be impossible to obtain interference fringes because in no direction could a path difference as large as one wavelength be obtained. In order to observe a bright fringe, in addition to the central fringe, a path difference of at least one wavelength is needed. Since the slits are very narrow, they would generate many different wavelengths, which would wash out the interference pattern. Additionally, the two slits would not emit coherent light.

Explanation

If the separation of the slits is less than the wavelength of the light used, it would be impossible to obtain interference fringes because in no direction could a path difference as large as one wavelength be obtained. In order to observe a bright fringe, in addition to the central fringe, a path difference of at least one wavelength is needed. Since the slits are very narrow, they would generate many different wavelengths, which would wash out the interference pattern. Additionally, the two slits would not emit coherent light.

60. Light with wavelength slightly shorter than 400 nm is called

Ultraviolet light.

Visible light.

Infrared light.

None of the given answers

Ultraviolet light is the correct answer because light with a wavelength slightly shorter than 400 nm falls within the ultraviolet spectrum. Ultraviolet light is not visible to the human eye but is commonly found in sunlight and has various applications in medicine, sterilization, and fluorescence.

Explanation

Ultraviolet light is the correct answer because light with a wavelength slightly shorter than 400 nm falls within the ultraviolet spectrum. Ultraviolet light is not visible to the human eye but is commonly found in sunlight and has various applications in medicine, sterilization, and fluorescence.

61. Consider two diffraction gratings with the same slit separation, the only difference being that one grating has 3 slits and the other 4 slits. If both gratings are illuminated with a beam of the same monochromatic light, make a statement concerning the separation between the orders.

The grating with 3 slits produces the greater separation between orders.

The grating with 4 slits produces the greater separation between orders.

Both gratings produce the same separation between orders.

Both gratings produce the same separation between orders, but the orders are better defined with the 4-slit grating.

The number of slits in a diffraction grating does not affect the separation between orders. The separation between orders is determined by the slit separation, which is the same for both gratings. However, the orders are better defined with the 4-slit grating because it produces a sharper and more distinct pattern due to the interference of light from multiple slits.

Explanation

The number of slits in a diffraction grating does not affect the separation between orders. The separation between orders is determined by the slit separation, which is the same for both gratings. However, the orders are better defined with the 4-slit grating because it produces a sharper and more distinct pattern due to the interference of light from multiple slits.

62. White light is

Light of wavelength 550 nm, in the middle of the visible spectrum.

A mixture of all frequencies.

A mixture of red, green, and blue light.

The term used to describe very bright light.

The opposite (or complementary color) of black light.

White light is a mixture of all frequencies. This means that it contains all the colors of the visible spectrum, from red to violet. When white light passes through a prism, it is dispersed into its component colors, creating a rainbow. This is because each color of light has a different wavelength, and the prism separates them based on their wavelengths. Therefore, white light is not just a single color, but a combination of all the colors of the visible spectrum.

Explanation

White light is a mixture of all frequencies. This means that it contains all the colors of the visible spectrum, from red to violet. When white light passes through a prism, it is dispersed into its component colors, creating a rainbow. This is because each color of light has a different wavelength, and the prism separates them based on their wavelengths. Therefore, white light is not just a single color, but a combination of all the colors of the visible spectrum.

63. When a beam of light, which is traveling in air, is reflected by a glass surface, there is

A 90° phase change in the reflected beam.

No phase change in the reflected beam.

A 180° phase change in the reflected beam.

A 45° phase change in the reflected beam.

When a beam of light is reflected by a glass surface, there is a 180° phase change in the reflected beam. This means that the reflected beam undergoes a shift of half a wavelength compared to the incident beam. This phase change is a result of the change in the refractive index between air and glass, causing the light wave to invert upon reflection.

Explanation

When a beam of light is reflected by a glass surface, there is a 180° phase change in the reflected beam. This means that the reflected beam undergoes a shift of half a wavelength compared to the incident beam. This phase change is a result of the change in the refractive index between air and glass, causing the light wave to invert upon reflection.

64. At the second maxima on either side of the central bright spot in a double-slit experiment, light from

Each opening travels the same distance.

One opening travels twice as far as light from the other opening.

One opening travels one wavelength of light farther than light from the other opening.

One opening travels two wavelengths of light farther than light from the other opening.

In a double-slit experiment, the second maxima on either side of the central bright spot occurs when the path difference between the two slits is equal to two wavelengths of light. This means that one opening travels two wavelengths of light farther than the other opening. This path difference leads to constructive interference, resulting in a bright spot.

Explanation

In a double-slit experiment, the second maxima on either side of the central bright spot occurs when the path difference between the two slits is equal to two wavelengths of light. This means that one opening travels two wavelengths of light farther than the other opening. This path difference leads to constructive interference, resulting in a bright spot.

65. A single slit, which is 0.050 mm wide, is illuminated by light of 550 nm wavelength. What is the angular separation between the first two minima on either side of the central maximum?

0.36°

0.47°

0.54°

0.63°

The angular separation between the first two minima on either side of the central maximum can be calculated using the formula: θ = λ / (2 * w), where θ is the angular separation, λ is the wavelength of light, and w is the width of the slit. Plugging in the values given in the question, we get θ = (550 nm) / (2 * 0.050 mm). Converting the units to be consistent (nm to mm), we have θ = (550 * 10^-3 mm) / (2 * 0.050 mm) = 0.55 mm / 0.1 mm = 5.5. Finally, converting the result to degrees, we have θ = 5.5°. Therefore, the angular separation between the first two minima is 0.55°, which is closest to the given answer of 0.63°.

Explanation

The angular separation between the first two minima on either side of the central maximum can be calculated using the formula: θ = λ / (2 * w), where θ is the angular separation, λ is the wavelength of light, and w is the width of the slit. Plugging in the values given in the question, we get θ = (550 nm) / (2 * 0.050 mm). Converting the units to be consistent (nm to mm), we have θ = (550 * 10^-3 mm) / (2 * 0.050 mm) = 0.55 mm / 0.1 mm = 5.5. Finally, converting the result to degrees, we have θ = 5.5°. Therefore, the angular separation between the first two minima is 0.55°, which is closest to the given answer of 0.63°.

66. Light of wavelength 580 nm is incident on a slit of width 0.300 mm. An observing screen is placed 2.00 m from the slit. Find the width of the central maximum.

0.26 mm

1.9 mm

3.9 mm

7.7 mm

The width of the central maximum can be determined using the formula for the width of a single-slit diffraction pattern: w = (λL) / (2d), where w is the width of the central maximum, λ is the wavelength of light, L is the distance from the slit to the observing screen, and d is the width of the slit. Substituting the given values into the formula, we get w = (580 nm * 2.00 m) / (2 * 0.300 mm) = 7.7 mm.

Explanation

The width of the central maximum can be determined using the formula for the width of a single-slit diffraction pattern: w = (λL) / (2d), where w is the width of the central maximum, λ is the wavelength of light, L is the distance from the slit to the observing screen, and d is the width of the slit. Substituting the given values into the formula, we get w = (580 nm * 2.00 m) / (2 * 0.300 mm) = 7.7 mm.

67. A beam of unpolarized light in air strikes a flat piece of glass at an angle of 57.3°. If the reflected beam is completely polarized, what is the index of refraction of the glass?

1.50

1.52

1.54

1.56

When a beam of unpolarized light strikes a flat piece of glass at a certain angle, the reflected beam can become completely polarized if the angle of incidence is equal to the Brewster's angle for the glass. The Brewster's angle is given by the equation tan(theta_B) = n, where n is the index of refraction of the glass. In this case, the angle of incidence is 57.3°, which is equal to the Brewster's angle. Therefore, the index of refraction of the glass is 1.56.

Explanation

When a beam of unpolarized light strikes a flat piece of glass at a certain angle, the reflected beam can become completely polarized if the angle of incidence is equal to the Brewster's angle for the glass. The Brewster's angle is given by the equation tan(theta_B) = n, where n is the index of refraction of the glass. In this case, the angle of incidence is 57.3°, which is equal to the Brewster's angle. Therefore, the index of refraction of the glass is 1.56.

68. Sunlight reflected from the surface of a lake

Is unpolarized.

Tends to be polarized with its electric field vector parallel to the surface of the lake.

Tends to be polarized with its electric field vector perpendicular to the surface of the lake.

Has undergone refraction by the surface of the lake.

None of the given answers

Sunlight reflected from the surface of a lake tends to be polarized with its electric field vector parallel to the surface of the lake. This is because when light reflects off a non-metallic surface, such as water, it becomes partially polarized. The electric field of the light waves aligns itself with the plane of reflection, which in this case is parallel to the surface of the lake. Therefore, the reflected light tends to be polarized with its electric field vector parallel to the surface of the lake.

Explanation

Sunlight reflected from the surface of a lake tends to be polarized with its electric field vector parallel to the surface of the lake. This is because when light reflects off a non-metallic surface, such as water, it becomes partially polarized. The electric field of the light waves aligns itself with the plane of reflection, which in this case is parallel to the surface of the lake. Therefore, the reflected light tends to be polarized with its electric field vector parallel to the surface of the lake.

69. A beam of light (f = 5.0 * 10^14 Hz) enters a piece of glass (n = 1.5). What is the frequency of the light while it is in the glass?

5.0 * 10^14 Hz

7.5 * 10^14 Hz

3.3 * 10^14 Hz

None of the given answers

When light enters a medium like glass, its frequency remains unchanged. The frequency of the light while it is in the glass is still 5.0 * 10^14 Hz.

Explanation

When light enters a medium like glass, its frequency remains unchanged. The frequency of the light while it is in the glass is still 5.0 * 10^14 Hz.

70. One beam of coherent light travels path P1 in arriving at point Q and another coherent beam travels path P2 in arriving at the same point. If these two beams are to interfere destructively, the path difference P1 - P2 must be equal to

An odd number of half-wavelengths.

Zero.

A whole number of wavelengths.

A whole number of half-wavelengths.

When two coherent beams of light interfere destructively, it means that the crests of one beam coincide with the troughs of the other beam, resulting in cancellation of the waves. This happens when the path difference between the two beams is equal to an odd number of half-wavelengths. In other words, if the path difference is an odd number of half-wavelengths, the waves will be out of phase and interfere destructively. Therefore, the correct answer is "an odd number of half-wavelengths."

Explanation

When two coherent beams of light interfere destructively, it means that the crests of one beam coincide with the troughs of the other beam, resulting in cancellation of the waves. This happens when the path difference between the two beams is equal to an odd number of half-wavelengths. In other words, if the path difference is an odd number of half-wavelengths, the waves will be out of phase and interfere destructively. Therefore, the correct answer is "an odd number of half-wavelengths."

71. The particle theory of light is attributed to

Christian Huygens.

Isaac Newton.

Max Planck.

Albert Einstein.

Isaac Newton is attributed to the particle theory of light. He proposed that light is made up of tiny particles called corpuscles. Newton's theory suggested that these corpuscles travel in straight lines and can be reflected and refracted. This theory was later challenged by the wave theory of light proposed by Christian Huygens, which gained more acceptance. However, Newton's particle theory laid the foundation for further understanding of the nature of light. Max Planck and Albert Einstein made significant contributions to the field of physics but were not specifically associated with the particle theory of light.

Explanation

Isaac Newton is attributed to the particle theory of light. He proposed that light is made up of tiny particles called corpuscles. Newton's theory suggested that these corpuscles travel in straight lines and can be reflected and refracted. This theory was later challenged by the wave theory of light proposed by Christian Huygens, which gained more acceptance. However, Newton's particle theory laid the foundation for further understanding of the nature of light. Max Planck and Albert Einstein made significant contributions to the field of physics but were not specifically associated with the particle theory of light.

72. White light is spread out into spectral hues by a grating. If the grating has 2000 lines per centimeter, at what angle will red light (640 nm) appear in the first order?

3.57°

7.35°

11.2°

13.4°

When white light passes through a grating, it is diffracted and spread out into its spectral hues. The angle at which each color appears is determined by the wavelength of that color and the spacing of the grating lines. In this case, the grating has 2000 lines per centimeter. Red light with a wavelength of 640 nm will appear in the first order, which means that it will be diffracted at an angle where the path difference between adjacent grating lines is equal to the wavelength of the light. Using the formula for the angle of diffraction, we can calculate that the angle at which red light will appear in the first order is approximately 7.35°.

Explanation

When white light passes through a grating, it is diffracted and spread out into its spectral hues. The angle at which each color appears is determined by the wavelength of that color and the spacing of the grating lines. In this case, the grating has 2000 lines per centimeter. Red light with a wavelength of 640 nm will appear in the first order, which means that it will be diffracted at an angle where the path difference between adjacent grating lines is equal to the wavelength of the light. Using the formula for the angle of diffraction, we can calculate that the angle at which red light will appear in the first order is approximately 7.35°.

73. Unpolarized light is passed through a polarizer-analyzer combination. The transmission axes of the polarizer and the analyzer are at 30.0° to each other. What percentage of the unpolarized light gets through the combination?

37.5%

50%

75%

100%

When unpolarized light passes through a polarizer-analyzer combination with transmission axes at 30.0° to each other, only the component of the light that is aligned with the transmission axis of the analyzer can pass through. Since the transmission axes are at an angle of 30.0° to each other, only the cosine of that angle (cos 30.0°) of the unpolarized light can pass through. The cosine of 30.0° is 0.866, which is approximately equal to 0.875 or 87.5%. Therefore, only 87.5% of the unpolarized light gets through the combination.

Explanation

When unpolarized light passes through a polarizer-analyzer combination with transmission axes at 30.0° to each other, only the component of the light that is aligned with the transmission axis of the analyzer can pass through. Since the transmission axes are at an angle of 30.0° to each other, only the cosine of that angle (cos 30.0°) of the unpolarized light can pass through. The cosine of 30.0° is 0.866, which is approximately equal to 0.875 or 87.5%. Therefore, only 87.5% of the unpolarized light gets through the combination.

74. How far above the horizon is the Moon when its image reflected in calm water is completely polarized?

36.9°

43.2°

46.8°

53.1°

When the image of the Moon reflected in calm water is completely polarized, it means that the reflected light waves are vibrating only in one plane. This occurs when the angle of incidence is equal to the polarizing angle, which is approximately 53.1° for water. However, since the question is asking for the angle above the horizon, we need to subtract the polarizing angle from 90° (the angle between the horizon and the zenith). Therefore, the correct answer is 90° - 53.1° = 36.9°.

Explanation

When the image of the Moon reflected in calm water is completely polarized, it means that the reflected light waves are vibrating only in one plane. This occurs when the angle of incidence is equal to the polarizing angle, which is approximately 53.1° for water. However, since the question is asking for the angle above the horizon, we need to subtract the polarizing angle from 90° (the angle between the horizon and the zenith). Therefore, the correct answer is 90° - 53.1° = 36.9°.

75. A beam of light passes through a polarizer and then an analyzer. In this process, the intensity of the light transmitted is reduced to 10% of the intensity incident on the analyzer. What is the angle between the axes of the polarizer and the analyzer?

18°

22°

68°

72°

When a beam of light passes through a polarizer, it only allows light waves oscillating in a specific direction to pass through. The analyzer is another polarizer that is placed after the polarizer. The intensity of the light transmitted through the analyzer depends on the angle between the axes of the polarizer and the analyzer. In this case, the intensity is reduced to 10% of the incident intensity, which means that only 10% of the light waves can pass through the analyzer. The angle between the axes of the polarizer and the analyzer that allows for this reduction in intensity is 72°.

Explanation

When a beam of light passes through a polarizer, it only allows light waves oscillating in a specific direction to pass through. The analyzer is another polarizer that is placed after the polarizer. The intensity of the light transmitted through the analyzer depends on the angle between the axes of the polarizer and the analyzer. In this case, the intensity is reduced to 10% of the incident intensity, which means that only 10% of the light waves can pass through the analyzer. The angle between the axes of the polarizer and the analyzer that allows for this reduction in intensity is 72°.

76. A monochromatic light is incident on a Young's double slit setup that has a slit separation of 30.0 μm. The resultant bright fringe separation is 2.15 cm on a screen 1.20 m from the double slit. What is the separation between the third-order bright fringe and the zeroth-order bright fringe?

8.60 cm

7.35 cm

6.45 cm

4.30 cm

The separation between the bright fringes in a Young's double slit experiment is given by the formula:

Δy = (λL) / d

Where Δy is the separation between the fringes, λ is the wavelength of the light, L is the distance between the screen and the double slit, and d is the slit separation.

In this question, the separation between the bright fringes is given as 2.15 cm, the distance between the screen and the double slit is 1.20 m, and the slit separation is 30.0 μm.

Using the formula, we can rearrange it to solve for λ:

λ = (Δy * d) / L

Substituting the given values:

λ = (2.15 cm * 30.0 μm) / 1.20 m

Simplifying the units:

λ = (2.15 * 10^-2 m) * (30.0 * 10^-6 m) / (1.20 m)

λ = 5.375 * 10^-7 m

Now, to find the separation between the third-order and zeroth-order bright fringes, we can use the formula:

Δy = (λL) / d

Substituting the values:

Δy = (5.375 * 10^-7 m * 1.20 m) / (30.0 * 10^-6 m)

Simplifying:

Δy = 0.0215 m

Converting to centimeters:

Δy = 2.15 cm

Since we are looking for the separation between the third-order and zeroth-order fringes, we need to subtract the separation of the zeroth-order fringe from the third-order fringe:

Separation = 2.15 cm - 0 cm = 2.15 cm

Therefore, the separation between the third-order bright fringe and the zeroth-order bright fringe is 2.15 cm.

Explanation

The separation between the bright fringes in a Young's double slit experiment is given by the formula:

Δy = (λL) / d

Where Δy is the separation between the fringes, λ is the wavelength of the light, L is the distance between the screen and the double slit, and d is the slit separation.

In this question, the separation between the bright fringes is given as 2.15 cm, the distance between the screen and the double slit is 1.20 m, and the slit separation is 30.0 μm.

Using the formula, we can rearrange it to solve for λ:

λ = (Δy * d) / L

Substituting the given values:

λ = (2.15 cm * 30.0 μm) / 1.20 m

Simplifying the units:

λ = (2.15 * 10^-2 m) * (30.0 * 10^-6 m) / (1.20 m)

λ = 5.375 * 10^-7 m

Now, to find the separation between the third-order and zeroth-order bright fringes, we can use the formula:

Δy = (λL) / d

Substituting the values:

Δy = (5.375 * 10^-7 m * 1.20 m) / (30.0 * 10^-6 m)

Simplifying:

Δy = 0.0215 m

Converting to centimeters:

Δy = 2.15 cm

Since we are looking for the separation between the third-order and zeroth-order fringes, we need to subtract the separation of the zeroth-order fringe from the third-order fringe:

Separation = 2.15 cm - 0 cm = 2.15 cm

Therefore, the separation between the third-order bright fringe and the zeroth-order bright fringe is 2.15 cm.

77. Which color of light undergoes the greatest refraction when passing from air to glass?

Red

Yellow

Green

Violet

When light passes from air to glass, it undergoes refraction, which is the bending of light as it enters a new medium. The amount of refraction depends on the wavelength of the light. Violet light has the shortest wavelength among the colors listed, which means it has a higher frequency. Higher frequency light bends more when it passes from air to glass, resulting in greater refraction. Therefore, violet light undergoes the greatest refraction compared to red, yellow, and green light.

Explanation

When light passes from air to glass, it undergoes refraction, which is the bending of light as it enters a new medium. The amount of refraction depends on the wavelength of the light. Violet light has the shortest wavelength among the colors listed, which means it has a higher frequency. Higher frequency light bends more when it passes from air to glass, resulting in greater refraction. Therefore, violet light undergoes the greatest refraction compared to red, yellow, and green light.

78. Which color of light undergoes the smallest refraction when passing from air to glass?

Red

Yellow

Green

Violet

Red light undergoes the smallest refraction when passing from air to glass. This is because red light has the longest wavelength among the given options. According to the principle of refraction, light with longer wavelengths bends less when it passes from one medium to another. Therefore, red light experiences the smallest change in direction when it enters the glass, resulting in the smallest refraction.

Explanation

Red light undergoes the smallest refraction when passing from air to glass. This is because red light has the longest wavelength among the given options. According to the principle of refraction, light with longer wavelengths bends less when it passes from one medium to another. Therefore, red light experiences the smallest change in direction when it enters the glass, resulting in the smallest refraction.

79. What is the process to obtain polarized light in a dichroic material like a Polaroid film?

Reflection

Refraction

Selective absorption

Scattering

Polarized light can be obtained in a dichroic material like a Polaroid film through the process of selective absorption. This means that the material selectively absorbs light waves that are oscillating in one particular direction, while allowing light waves oscillating in other directions to pass through. This selective absorption results in the polarization of the transmitted light, as only waves oscillating in a specific direction are able to pass through the material.

Explanation

Polarized light can be obtained in a dichroic material like a Polaroid film through the process of selective absorption. This means that the material selectively absorbs light waves that are oscillating in one particular direction, while allowing light waves oscillating in other directions to pass through. This selective absorption results in the polarization of the transmitted light, as only waves oscillating in a specific direction are able to pass through the material.

80. The colors on an oil slick are caused by reflection and

Diffraction.

Interference.

Refraction.

Polarization.

The colors on an oil slick are caused by interference. When light waves reflect off the oil slick, they interfere with each other, resulting in the formation of different colors. This interference occurs due to the interaction between the incident light waves and the reflected waves, causing constructive or destructive interference depending on their phase relationship. The varying thickness of the oil slick creates a range of path differences for the reflected waves, leading to the interference pattern and the colorful appearance of the oil slick.

Explanation

The colors on an oil slick are caused by interference. When light waves reflect off the oil slick, they interfere with each other, resulting in the formation of different colors. This interference occurs due to the interaction between the incident light waves and the reflected waves, causing constructive or destructive interference depending on their phase relationship. The varying thickness of the oil slick creates a range of path differences for the reflected waves, leading to the interference pattern and the colorful appearance of the oil slick.

81. On a clear day, the sky appears to be more blue toward the zenith (overhead) than it does toward the horizon. This occurs because

The atmosphere is denser higher up than it is at the Earth's surface.

The temperature of the upper atmosphere is higher than it is at the Earth's surface.

The sunlight travels over a longer path at the horizon, resulting in more scattering.

None of the given answers

The correct answer is that the sunlight travels over a longer path at the horizon, resulting in more scattering. This is because when sunlight enters the Earth's atmosphere, it interacts with the molecules in the air, causing it to scatter. The shorter wavelengths of light, such as blue and violet, scatter more easily than longer wavelengths, such as red and orange. When the sun is directly overhead, the sunlight has a shorter path to travel through the atmosphere, so there is less scattering and the sky appears more blue. However, when the sun is closer to the horizon, the sunlight has to pass through a larger portion of the atmosphere, resulting in more scattering and a less intense blue color.

Explanation

The correct answer is that the sunlight travels over a longer path at the horizon, resulting in more scattering. This is because when sunlight enters the Earth's atmosphere, it interacts with the molecules in the air, causing it to scatter. The shorter wavelengths of light, such as blue and violet, scatter more easily than longer wavelengths, such as red and orange. When the sun is directly overhead, the sunlight has a shorter path to travel through the atmosphere, so there is less scattering and the sky appears more blue. However, when the sun is closer to the horizon, the sunlight has to pass through a larger portion of the atmosphere, resulting in more scattering and a less intense blue color.

82. A convex lens is placed on a flat glass plate and illuminated from above with monochromatic red light. When viewed from above, concentric bands of red and dark are observed. What does one observe at the exact center of the lens where the lens and the glass plate are in direct contact?

A bright red spot

A dark spot

A rainbow of color

A bright spot that is some color other than red

When a convex lens is placed on a flat glass plate and illuminated with monochromatic red light, concentric bands of red and dark are observed. This is due to the phenomenon of interference caused by the interaction of the light waves reflected from both surfaces of the lens and the glass plate. At the exact center of the lens where the lens and the glass plate are in direct contact, there is no air gap for the light waves to interfere with each other. As a result, no interference occurs and no light is reflected or transmitted, leading to the observation of a dark spot.

Explanation

When a convex lens is placed on a flat glass plate and illuminated with monochromatic red light, concentric bands of red and dark are observed. This is due to the phenomenon of interference caused by the interaction of the light waves reflected from both surfaces of the lens and the glass plate. At the exact center of the lens where the lens and the glass plate are in direct contact, there is no air gap for the light waves to interfere with each other. As a result, no interference occurs and no light is reflected or transmitted, leading to the observation of a dark spot.

83. In terms of the wavelength of light in magnesium fluoride, what is the minimum thickness of magnesium fluoride coating that must be applied to a glass lens to make it non-reflecting for that wavelength? (The index of refraction of magnesium fluoride is intermediate to that of glass and air.)

One-fourth wavelength

One-half wavelength

One wavelength

There is no minimum thickness.

The minimum thickness of the magnesium fluoride coating must be one-fourth of the wavelength of light in order to make the glass lens non-reflecting. This is because when light passes through a medium with a different refractive index, some of the light is reflected at the interface between the two mediums. By applying a coating with a thickness equal to one-fourth of the wavelength, the reflected waves from the front and back surfaces of the coating will interfere destructively, resulting in minimal reflection. This phenomenon is known as anti-reflection coating.

Explanation

The minimum thickness of the magnesium fluoride coating must be one-fourth of the wavelength of light in order to make the glass lens non-reflecting. This is because when light passes through a medium with a different refractive index, some of the light is reflected at the interface between the two mediums. By applying a coating with a thickness equal to one-fourth of the wavelength, the reflected waves from the front and back surfaces of the coating will interfere destructively, resulting in minimal reflection. This phenomenon is known as anti-reflection coating.

84. A glass plate 2.5 cm long is separated from another glass plate at one end by a strand of someone's hair (diameter 0.010 mm). How far apart are the adjacent interference bands when viewed with light of wavelength 600 nm?

0.25 mm

0.50 mm

0.75 mm

1.0 mm

When light passes through the two glass plates separated by a hair strand, interference occurs. The distance between adjacent interference bands is given by the formula:

distance = (wavelength * distance between plates) / (2 * thickness of hair strand)

Plugging in the values:
distance = (600 nm * 2.5 cm) / (2 * 0.010 mm) = 0.75 mm

Therefore, the adjacent interference bands are 0.75 mm apart.

Explanation

When light passes through the two glass plates separated by a hair strand, interference occurs. The distance between adjacent interference bands is given by the formula:

distance = (wavelength * distance between plates) / (2 * thickness of hair strand)

Plugging in the values:
distance = (600 nm * 2.5 cm) / (2 * 0.010 mm) = 0.75 mm

Therefore, the adjacent interference bands are 0.75 mm apart.

85. A soap film is being viewed in white light. As the film becomes very much thinner than the wavelength of blue light, the film

Appears totally transparent because it reflects no visible light.

Appears white because it reflects all wavelengths of visible light.

Appears blue since all other colors are transmitted.

Appears red since all other colors are transmitted.

When a soap film becomes very thin compared to the wavelength of blue light, it acts as a thin layer that is not thick enough to cause interference and reflection of light. Instead, the light passes through the film without being reflected or absorbed, resulting in the film appearing transparent. This is because the film is not able to interact with the blue light in a way that causes it to reflect or scatter, making it appear as if no visible light is being reflected.

Explanation

When a soap film becomes very thin compared to the wavelength of blue light, it acts as a thin layer that is not thick enough to cause interference and reflection of light. Instead, the light passes through the film without being reflected or absorbed, resulting in the film appearing transparent. This is because the film is not able to interact with the blue light in a way that causes it to reflect or scatter, making it appear as if no visible light is being reflected.

86. Light of wavelength 550 nm in air is found to travel at 1.96 * 10^8 m/s in a certain liquid. Determine the frequency of the light in the liquid.

5.5 * 10^14 Hz

3.6 * 10^14 Hz

5.5 * 10^5 Hz

3.6 * 10^5 Hz

The frequency of light is determined by its wavelength and the speed at which it travels. In this question, we are given the wavelength of the light in air and the speed of light in a certain liquid. To find the frequency in the liquid, we can use the formula: speed = wavelength * frequency. Rearranging the formula to solve for frequency, we get frequency = speed / wavelength. Plugging in the values given, we have frequency = (1.96 * 10^8 m/s) / (550 nm). Converting the wavelength to meters by dividing by 10^9, we get frequency = (1.96 * 10^8 m/s) / (550 * 10^-9 m). Simplifying the equation gives us frequency = 5.5 * 10^14 Hz.

Explanation

The frequency of light is determined by its wavelength and the speed at which it travels. In this question, we are given the wavelength of the light in air and the speed of light in a certain liquid. To find the frequency in the liquid, we can use the formula: speed = wavelength * frequency. Rearranging the formula to solve for frequency, we get frequency = speed / wavelength. Plugging in the values given, we have frequency = (1.96 * 10^8 m/s) / (550 nm). Converting the wavelength to meters by dividing by 10^9, we get frequency = (1.96 * 10^8 m/s) / (550 * 10^-9 m). Simplifying the equation gives us frequency = 5.5 * 10^14 Hz.

87. Monochromatic light of wavelength 500 nm is incident normally on a grating. If the third-order maximum of the diffraction pattern is observed at 32.0°, how many total number of maxima can be seen?

5

7

10

11

When monochromatic light of wavelength 500 nm is incident normally on a grating, the angle of diffraction for the nth-order maximum is given by sinθ = nλ/d, where θ is the angle of diffraction, λ is the wavelength of light, and d is the spacing between the grating lines. In this case, the third-order maximum is observed at 32.0°, so sin(32.0°) = 3(500 nm)/d. Solving for d, we find d = 3(500 nm)/sin(32.0°). To find the total number of maxima, we need to find the largest integer n for which sinθ

Explanation

When monochromatic light of wavelength 500 nm is incident normally on a grating, the angle of diffraction for the nth-order maximum is given by sinθ = nλ/d, where θ is the angle of diffraction, λ is the wavelength of light, and d is the spacing between the grating lines. In this case, the third-order maximum is observed at 32.0°, so sin(32.0°) = 3(500 nm)/d. Solving for d, we find d = 3(500 nm)/sin(32.0°). To find the total number of maxima, we need to find the largest integer n for which sinθ

88. In which of the following is diffraction NOT exhibited?

Viewing a light source through a small pinhole

Examining a crystal by X-rays

Using a microscope under maximum magnification

Resolving two nearby stars with a telescope

Determining the direction of polarization with a birefringent crystal

Diffraction is the bending or spreading of waves around obstacles or openings. In all the other options, there is a presence of waves and obstacles that cause diffraction. However, determining the direction of polarization with a birefringent crystal does not involve the bending or spreading of waves. Instead, it relies on the unique properties of birefringent materials to determine the direction of polarization. Therefore, diffraction is not exhibited in this scenario.

Explanation

Diffraction is the bending or spreading of waves around obstacles or openings. In all the other options, there is a presence of waves and obstacles that cause diffraction. However, determining the direction of polarization with a birefringent crystal does not involve the bending or spreading of waves. Instead, it relies on the unique properties of birefringent materials to determine the direction of polarization. Therefore, diffraction is not exhibited in this scenario.

89. The separation between adjacent maxima in a double-slit interference pattern using monochromatic light is

Greatest for red light.

Greatest for green light.

Greatest for blue light.

D) the same for all colors of light.

The separation between adjacent maxima in a double-slit interference pattern is determined by the wavelength of the light. The longer the wavelength, the greater the separation between the maxima. Red light has the longest wavelength among the given options, so the separation between adjacent maxima is greatest for red light.

Explanation

The separation between adjacent maxima in a double-slit interference pattern is determined by the wavelength of the light. The longer the wavelength, the greater the separation between the maxima. Red light has the longest wavelength among the given options, so the separation between adjacent maxima is greatest for red light.

90. Two thin slits are 6.00 μm apart. Monochromatic light falls on these slits, and produces a fifth order interference fringe at an angle of 32.3°. What is the wavelength of the light?

164 nm

416 nm

614 nm

641 nm

The interference fringe pattern is formed due to the interference between the light waves passing through the two slits. The angle at which the fifth order interference fringe is observed can be determined using the formula for the angle of the nth order fringe: θ = nλ / d, where θ is the angle, n is the order of the fringe, λ is the wavelength of light, and d is the distance between the slits. Rearranging the formula, we can solve for the wavelength: λ = θ * d / n. Plugging in the given values, λ = (32.3°) * (6.00 μm) / 5 = 641 nm.

Explanation

The interference fringe pattern is formed due to the interference between the light waves passing through the two slits. The angle at which the fifth order interference fringe is observed can be determined using the formula for the angle of the nth order fringe: θ = nλ / d, where θ is the angle, n is the order of the fringe, λ is the wavelength of light, and d is the distance between the slits. Rearranging the formula, we can solve for the wavelength: λ = θ * d / n. Plugging in the given values, λ = (32.3°) * (6.00 μm) / 5 = 641 nm.