# Physics Quiz: Light And Optics

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| Written by Junho Song
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Junho Song, Physics/Science Teacher
Junho Song, a dedicated educator, specializes in teaching physics and science. Passionate about inspiring students in the fascinating world of scientific discovery and understanding.
Quizzes Created: 11 | Total Attempts: 80,141
Questions: 15 | Attempts: 18,035  Settings  Hello and welcome to this ultimate Physics quiz on Light and Optics that we've created below for you. This quiz will test your knowledge about light and its related concepts. Let's see if you can crack this quiz or not. The questions here are multiple-choice based, and you just have to do is choose the correct answer to every question. Do you think you think you'll be able to pass this test? Let's start the quiz and find out.

• 1.

### A survivor from a shipwreck sees an image of a fish in the water. To catch it with her spear, she must:

• A.

Aim above the image of the fish

• B.

Aim below the image of the fish

• C.

Aim at the image of the fish

• D.

Aim behind the fish

B. Aim below the image of the fish
Explanation
When light travels from one medium to another, it changes direction. This phenomenon is called refraction. When the survivor sees the image of the fish in the water, she is actually seeing a refracted image due to the bending of light. The image appears higher than the actual position of the fish. Therefore, to compensate for this refraction, she needs to aim below the image of the fish in order to accurately spear it.

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• 2.

### A light ray has an angle of incidence of 34º. The reflected ray will make what angle with the reflecting surface?

• A.

• B.

34º

• C.

66º

• D.

74º

B. 34º
Explanation
When a light ray is incident on a reflecting surface, the angle of incidence is equal to the angle of reflection. In this case, the angle of incidence is 34°, so the reflected ray will also make an angle of 34° with the reflecting surface.

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• 3.

### The critical angle for diamond (n = 2.42) submerged in water (n = 1.33) is:

• A.

33.34°

• B.

49°

• C.

24°

• D.

17°

• E.

Does not exist

A. 33.34°
Explanation
The critical angle for diamond submerged in water can be calculated using the formula: critical angle = sin^(-1)(n2/n1), where n2 is the refractive index of the medium diamond is submerged in (water) and n1 is the refractive index of diamond. Plugging in the values, we get critical angle = sin^(-1)(1.33/2.42) = 33.34°. Therefore, the correct answer is 33.34°.

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• 4.

### Calculate the index of refraction for an object in which light travels at 1.97 x 108 m/s.

• A.

1.97 m/s

• B.

1.52 m/s

• C.

1.95

• D.

1.52

D. 1.52
Explanation
The index of refraction is a measure of how much light slows down when it passes through a medium compared to its speed in a vacuum. In this question, the given speed of light in the object is 1.97 x 10^8 m/s. To calculate the index of refraction, we need to divide the speed of light in a vacuum (3 x 10^8 m/s) by the speed of light in the object. Therefore, the index of refraction is 3 x 10^8 m/s divided by 1.97 x 10^8 m/s, which equals approximately 1.52.

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• 5.

### The critical angle of zircon is 31°. Which of the following incident angles would result in total internal reflection?

• A.

17°

• B.

34°

• C.

56°

• D.

B and C

D. B and C
Explanation
The critical angle is the angle of incidence at which light passing from a denser medium to a less dense medium is refracted at an angle of 90 degrees. Any incident angle greater than the critical angle will result in total internal reflection. In this case, the critical angle of zircon is 31 degrees. Therefore, incident angles of 34 degrees and 56 degrees (options B and C) would result in total internal reflection.

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• 6.

### When a light ray traveling in glass is incident on an air surface:

• A.

It will refract away from the normal

• B.

Some of the light may be reflected

• C.

All of the light may be reflected

• D.

All of A, B, and C

• E.

Two of A, B, and C

D. All of A, B, and C
Explanation
When a light ray traveling in glass is incident on an air surface, it can undergo multiple phenomena. Firstly, it will refract away from the normal, meaning it will change direction as it passes from a denser medium (glass) to a less dense medium (air). Additionally, some of the light may be reflected at the air surface, causing a partial reflection. Lastly, under certain conditions, all of the light may be reflected, resulting in total internal reflection. Therefore, all of the options A, B, and C are correct, as they describe different possibilities that can occur when a light ray travels from glass to air.

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• 7.

### Light travels from medium X into medium Y. Medium Y has a higher index of refraction. Consider each statement below: (i) The light travels faster in X. (ii) The light will bend towards the normal. (iii) The light will speed up. (iv) The light will bend away from the normal.

• A.

(ii)

• B.

(iii) and (iv)

• C.

(i) and (ii)

• D.

(ii), (iii), and (iv)

C. (i) and (ii)
Explanation
When light travels from a medium with a lower index of refraction to a medium with a higher index of refraction, it slows down and bends towards the normal. This is because the speed of light is inversely proportional to the index of refraction of the medium it is traveling in. Therefore, statement (i) is correct because the light travels slower in medium X compared to medium Y. Statement (ii) is also correct because the light will indeed bend towards the normal when it enters medium Y.

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• 8.

### For a converging lens, a light ray that is traveling parallel to the principal axis refracts:

• A.

Through the principal focus

• B.

Through the secondary focus

• C.

Through the optical centre

• D.

Parallel to the principal axis

A. Through the principal focus
Explanation
A converging lens is designed to bring parallel rays of light to a single point called the principal focus. When a light ray travels parallel to the principal axis and encounters a converging lens, it refracts and passes through the lens in such a way that it converges to a point at the principal focus. Therefore, the correct answer is "through the principal focus."

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• 9.

### A object is placed between f and 2f for a diverging lens. The image will be located:

• A.

Between f and 2f

• B.

Between the lens and f

• C.

Farther than 2f

• D.

A or B

B. Between the lens and f
Explanation
When an object is placed between the focal point (f) and twice the focal point (2f) for a diverging lens, the image formed will be located between the lens and the focal point (f). This is because a diverging lens always forms virtual and upright images that are located on the same side as the object. Since the object is placed between f and 2f, the image will be formed on the same side of the lens, between the lens and f.

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• 10.

### Light that travels into the eye passes through several parts to get to the retina. The correct order is:

• A.

Cornea, vitreous humour, lens, pupil

• B.

Cornea, lens, pupil, vitreous humour

• C.

Pupil, cornea, lens, vitreous humour

• D.

Cornea, pupil, lens, vitreous humour

D. Cornea, pupil, lens, vitreous humour
Explanation
Light enters the eye through the cornea, which is the clear, outermost layer of the eye. It then passes through the pupil, which is the opening in the center of the iris that regulates the amount of light entering the eye. After passing through the pupil, the light reaches the lens, which focuses the light onto the retina at the back of the eye. Finally, the light passes through the vitreous humour, a gel-like substance that fills the space between the lens and the retina, before reaching the retina where it is converted into electrical signals for the brain to process.

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• 11.

### The proper name for the inability to focus in different planes is:

• A.

Hyperopia

• B.

Presbyopia

• C.

Astigmatism

• D.

Myopia

C. Astigmatism
Explanation
Astigmatism is the proper name for the inability to focus in different planes. It is a common eye condition where the cornea or lens is irregularly shaped, causing blurry or distorted vision at all distances. This irregularity prevents light from properly focusing on the retina, resulting in difficulties in focusing on objects both near and far. Hyperopia refers to farsightedness, presbyopia is the loss of near vision with age, and myopia is nearsightedness.

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• 12.

### The proper name for farsightedness is:

• A.

Hyperopia

• B.

Presbyopia

• C.

Astigmatism

• D.

Glaucoma

A. Hyperopia
Explanation
Hyperopia, also known as farsightedness, is a common refractive error where distant objects are seen more clearly than objects that are near. It occurs when the eyeball is shorter than normal or when the cornea is too flat, causing light entering the eye to focus behind the retina instead of directly on it. This results in blurred vision for close objects. Presbyopia, on the other hand, is a condition that occurs with age and affects the eye's ability to focus on close objects. Astigmatism and glaucoma are different eye conditions unrelated to farsightedness.

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• 13.

### A block of glass is pushed into the path of the light, as shown below. The point where the light rays cross will:

• A.

Stay in the same place

• B.

Move to the left

• C.

Move to the right

• D.

Shift up

• E.

Shift down

C. Move to the right
Explanation
When a block of glass is pushed into the path of light, it causes the light rays to change direction due to refraction. Refraction occurs because the speed of light changes when it passes from one medium (air) to another (glass). In this case, the block of glass is denser than air, so the light rays bend towards the normal (a line perpendicular to the surface of the glass). As a result, the point where the light rays cross will appear to move to the right.

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• 14.

### The focal length of a converging lens is 15 cm. An object is placed 45 cm away from the lens. The image will be:

• A.

Smaller and real

• B.

Larger and real

• C.

The same size and real

• D.

Smaller and virtual

• E.

Larger and virtual

A. Smaller and real
Explanation
When an object is placed beyond the focal point of a converging lens, a real and smaller image is formed on the opposite side of the lens. This is because the light rays converge after passing through the lens, creating a smaller image. Therefore, the given answer "smaller and real" is correct.

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• 15.

### The focal length of a diverging lens is 12 cm. An object is placed 5.0 cm away from the lens. The image will be:

• A.

Smaller and real

• B.

Larger and real

• C.

The same size and real

• D.

Smaller and virtual Back to top