# Physics Quiz: Trivia Questions On Wave Optics!

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| Written by Tanmay Shankar
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Tanmay Shankar
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Quizzes Created: 547 | Total Attempts: 1,801,734
Questions: 15 | Attempts: 679  Settings  .

• 1.

### For sustained interference, we need two sources which emit radiations:

• A.

Of the same intensity

• B.

Of the same amplitude

• C.

Having a constant phase difference

• D.

None of these

C. Having a constant phase difference
Explanation
In order to observe sustained interference, it is necessary for two sources to emit radiations with a constant phase difference. This means that the two sources are emitting waves that are in sync with each other, with a consistent phase relationship. If the phase difference between the waves is not constant, the interference pattern will not be sustained and will fluctuate over time. Therefore, it is crucial for the phase difference to remain constant for sustained interference to occur.

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• 2.

• A.

2I

• B.

4I

• C.

5I

• D.

7I

B. 4I
• 3.

### n identical waves each of intensity Io interfere each other. The ratio of maximum intensities if interference is coherent and incoherent is:

• A.

N

• B.

N2

• C.

N3

• D.

1/n2

A. N
Explanation
When n identical waves interfere coherently, the maximum intensity is obtained when all the waves are in phase and add up constructively. In this case, the intensity is directly proportional to the number of waves, so the ratio of maximum intensities would be n.

On the other hand, when the interference is incoherent, the waves do not maintain a constant phase relationship. As a result, the waves do not add up perfectly and the maximum intensity is less than the coherent case. However, the ratio of maximum intensities still depends on the number of waves, so the answer remains n.

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• 4.

### The initial shape of the wavefront of the beam is:

• A.

Concave

• B.

Convex near the axis and concave near the periphery

• C.

Planar

• D.

Convex

C. Planar
Explanation
The initial shape of the wavefront of the beam is planar, meaning that it is flat and does not curve in any direction. This suggests that the beam is propagating in a straight line without any distortion or curvature in its wavefront.

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• 5.

• A.
• B.
• C.
• D.
D.
• 6.

### The fringe width β of a diffraction pattern and the slit width d are related as:

• A.

βαd

• B.
• C.
• D.
B.
Explanation
The fringe width β of a diffraction pattern is directly proportional to the slit width d. This means that as the slit width increases, the fringe width also increases. This relationship can be understood by considering that a wider slit allows more light to pass through, resulting in a larger diffraction pattern with wider fringes. Conversely, a narrower slit restricts the amount of light passing through, leading to a smaller diffraction pattern with narrower fringes. Therefore, the fringe width and slit width are related, with the fringe width increasing as the slit width increases.

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• 7.

### Two periodic waves of intensities I1 and I2 pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is:

• A.
• B.
• C.
• D.
D.
Explanation
The sum of the maximum and minimum intensities of the two periodic waves is equal to the sum of their individual maximum and minimum intensities. This is because the waves are passing through the region at the same time and in the same direction, so their intensities add up. Therefore, the sum of the maximum and minimum intensities is I1(max) + I2(max) + I1(min) + I2(min).

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• 8.

### A beam is used in young’s double slit experiment. The slit width is d. When the velocity of electron is increased, then

• A.

No interference in observed

• B.

Fringe width increases

• C.

Fringe width decreases

• D.

Fringe width remains same

C. Fringe width decreases
Explanation
When the velocity of the electron is increased in Young's double slit experiment, the fringe width decreases. This is because the fringe width is inversely proportional to the velocity of the particle. As the velocity increases, the distance traveled by the electron in a given time interval also increases. This results in a shorter wavelength and a smaller fringe width. Therefore, the correct answer is "Fringe width decreases."

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• 9.

### A Polaroid produces a strong beam of light which is:

• A.

Circularly polarised

• B.

Elliptically polarised

• C.

Plane polarised

• D.

Unpolarised

C. Plane polarised
Explanation
A Polaroid produces a strong beam of light that is plane polarised. This means that the light waves oscillate in a single plane as they travel. Plane polarised light is created by passing unpolarised light through a polarising filter, such as a Polaroid sheet. The filter allows only the light waves vibrating in a specific direction to pass through, blocking the waves vibrating in other directions. As a result, the light that emerges from a Polaroid is plane polarised, with all the waves oscillating in the same plane.

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• 10.

### The angle between pass axis of polarizer and analyzer is 45o. The percentage of polarised light passing through analyzer is:

• A.

100%

• B.

50%

• C.

25%

• D.

75%

B. 50%
Explanation
When the angle between the pass axis of the polarizer and analyzer is 45 degrees, the intensity of the polarized light passing through the analyzer is reduced by half. This is because the analyzer only allows light that is polarized in the same direction as its pass axis to pass through, while blocking light polarized perpendicular to its pass axis. Therefore, only 50% of the polarized light will pass through the analyzer.

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• 11.

### A phase difference of 5π corresponds to a path difference (in terms λ) of:

• A.

• B.

10λ

• C.

5λ/2

• D.

C. 5λ/2
Explanation
A phase difference of 5π corresponds to a path difference (in terms of λ) of 5λ/2. This can be explained by understanding the relationship between phase difference and path difference in wave interference. In wave interference, the phase difference between two waves is directly related to the path difference between them. When the phase difference is 2π, the path difference is equal to one wavelength (λ). Therefore, when the phase difference is 5π, the path difference would be 5/2 times the wavelength (5λ/2).

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• 12.

### Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are:

• A.

5I and I

• B.

9I and I

• C.

5I and 3I

• D.

9I and 3I

B. 9I and I
Explanation
When two coherent monochromatic light beams are superposed, the resulting intensity can be calculated using the principle of superposition. The principle states that the intensities of the two beams add up algebraically. In this case, the intensities are I and 4I. Adding these together gives a maximum intensity of 5I (when the two beams are in phase) and a minimum intensity of I (when the two beams are completely out of phase). Therefore, the correct answer is 5I and I.

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• 13.

### What happens if one of the slits, say S1, in Young’s double slit experiment is covered with a glass plate which absorbs half the intensity of light from it?

• A.

The bright fringes become less bright and dark fringes have a finite light intensity

• B.

The bright fringes become brighter and dark fringes become darker

• C.

The fringe width decreases

• D.

No fringes will be observed

A. The bright fringes become less bright and dark fringes have a finite light intensity
Explanation
When one of the slits, S1, is covered with a glass plate that absorbs half the intensity of light from it, the overall intensity of light reaching the screen decreases. This causes the bright fringes to become less bright as there is less light contributing to their intensity. Additionally, the dark fringes have a finite light intensity because even though some light is being absorbed by the glass plate, there is still some light passing through the other slit, S2, and reaching the dark regions. Therefore, the bright fringes become less bright and the dark fringes have a finite light intensity.

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• 14.

### In a double slit experiment, the distance between the slits is d. The screen is at a distance D from the slits. If a bright fringe is formed opposite of one of the slits, find its order:

• A.
• B.
• C.
• D.
D.
Explanation
The order of the bright fringe can be determined using the equation mλ = d sinθ, where m is the order of the fringe, λ is the wavelength of light, d is the distance between the slits, and θ is the angle between the line joining the slit and the line joining the slit and the bright fringe. Since the bright fringe is formed opposite one of the slits, the angle θ is 0°. Therefore, sinθ is also 0, and the equation becomes mλ = 0. This implies that the order of the bright fringe is 0.

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• 15.

### Two coherent monochromatic light sources are located at two vertices of an equilateral triangle. If the intensity due to each of the sources independently is 1Wm-2 at the third vertex, the resultant intensity due to both the sources at that point (i.e., at the third vertex) is:

• A.

Zero

• B.
• C.

2

• D.

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