College Physics Quiz: Refraction

15 questions · exam conditions

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RefractionQuestion 1 of 15

A ray of light traveling in glass (n = 1.50) approaches the glass-air interface. At what minimum angle of incidence will the ray undergo total internal reflection rather than partial transmission into air?

30.0°

41.8°

48.6°

60.0°

66.9°

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College Physics Quiz

College Physics Quiz: Refraction

Practice Refraction in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Refraction, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

30.0°
41.8° (correct answer)
48.6°
60.0°
66.9°

Explanation: When light travels from a denser medium (glass) to a less dense medium (air), total internal reflection occurs when the angle of incidence exceeds the critical angle. This happens because beyond this critical angle, no light can be transmitted into the second medium—it's all reflected back. To find the critical angle, you use Snell's law at the point where the refracted ray would graze along the interface (refraction angle = 90°):

n_1 \sin θ_c = n_2 \sin 90°

Since

\sin 90° = 1

, this becomes:

\sin θ_c = \frac{n_2}{n_1} = \frac{1.00}{1.50} = 0.667

Therefore:

θ_c = \arcsin(0.667) = 41.8°

Choice B (41.8°) is correct—this is the minimum angle for total internal reflection. Choice A (30.0°) corresponds to

\arcsin(0.5)

, which you'd get if you incorrectly used

n_{air} = 0.75

instead of 1.00. Choice C (48.6°) is what you'd calculate if you mistakenly flipped the refractive indices:

\arcsin(1.50/1.00)

, but since this gives a value greater than 1, it's physically impossible. Choice D (60.0°) corresponds to

\arcsin(\sqrt{3}/2)

, suggesting confusion with common trigonometric values rather than proper application of Snell's law. Remember: for critical angle problems, the less dense medium always goes in the numerator, and the angle you calculate is the minimum angle for total internal reflection to begin.

Question 2

Two identical glass blocks are placed in contact, with a thin film of oil (n = 1.45) between them. The glass has a refractive index of 1.52. A light ray traveling in the glass strikes the glass-oil interface at 60° from the normal. What happens to the ray?

It undergoes total internal reflection since the critical angle is exceeded
It refracts into the oil at an angle of 65° from the normal (correct answer)
It refracts into the oil at an angle of 72° from the normal
It continues straight through without bending since both blocks are identical
It splits into reflected and transmitted components with the transmitted ray at 54° from the normal

Explanation: When light travels from one medium to another with different refractive indices, you need to apply Snell's law and check whether total internal reflection occurs. The key is comparing the refractive indices: light going from a denser medium (higher n) to a less dense medium (lower n) can undergo total internal reflection if the angle exceeds the critical angle. Here, light travels from glass (n = 1.52) to oil (n = 1.45), so it's going from denser to less dense. First, let's find the critical angle:

\sin θ_c = \frac{n_{oil}}{n_{glass}} = \frac{1.45}{1.52} = 0.954

, giving

θ_c = 72.4°

. Since the incident angle (60°) is less than the critical angle, the light will refract, not reflect. Using Snell's law:

n_{glass} \sin(60°) = n_{oil} \sin(θ_r)

1.52 × 0.866 = 1.45 × \sin(θ_r)

\sin(θ_r) = \frac{1.316}{1.45} = 0.907

θ_r = 65°

Answer A is wrong because 60° < 72.4° (the critical angle), so total internal reflection doesn't occur. Answer C incorrectly gives 72°, which might come from confusing the critical angle with the refraction angle. Answer D is wrong because even though the glass blocks are identical, the light must pass through the oil layer between them, and the oil has a different refractive index than glass. Always calculate the critical angle first when light goes from high to low refractive index—this tells you whether to use Snell's law or expect total internal reflection.

Question 3

A glass sphere (n = 1.50) is immersed in water (n = 1.33). A light ray in the water approaches the sphere and enters at an angle of incidence of 40° to the normal at the point of contact. At what angle does the ray emerge from the sphere back into the water on the opposite side?

30°
35°
40° (correct answer)
45°
50°

Explanation: When light travels between materials with different refractive indices, you need to apply Snell's law at each interface. This problem involves two refractions: water-to-glass and glass-to-water. First, let's find the refraction angle when light enters the glass sphere. Using Snell's law:

n_1 \sin \theta_1 = n_2 \sin \theta_2

For the water-to-glass interface:

1.33 \sin(40°) = 1.50 \sin \theta_2

\sin \theta_2 = \frac{1.33 \times 0.643}{1.50} = 0.570

\theta_2 = 34.7°

Now here's the key insight: due to the spherical symmetry of the glass sphere, when the ray reaches the opposite side to exit back into water, it hits that interface at the same angle to the normal as it traveled through the glass (34.7°). This is because radii of a sphere are parallel at opposite points. For the glass-to-water interface:

1.50 \sin(34.7°) = 1.33 \sin \theta_3

\sin \theta_3 = \frac{1.50 \times 0.570}{1.33} = 0.643

\theta_3 = 40°

The ray emerges at 40°, making C correct. A) 30° assumes incorrect refraction calculations. B) 35° might result from using the intermediate angle inside the glass as the final answer. D) 45° could come from assuming total internal reflection occurs, which it doesn't at these angles. Remember: for spherical optical elements, the symmetry often means that rays entering and exiting at corresponding points follow reversible paths, especially when the surrounding medium is the same.

Question 4

A student observes that when light travels from air into a certain plastic material at an angle of 50° from the normal, the refracted ray makes an angle of 30° with the normal. If the student then sends light from this plastic into water (n = 1.33), what angle of incidence in the plastic would produce an angle of refraction of 45° in the water?

28°
32°
38° (correct answer)
42°
46°

Explanation: When you encounter refraction problems involving multiple materials, you need to apply Snell's law systematically and find the refractive index of each material first. Start by finding the plastic's refractive index using the air-to-plastic data. Snell's law states

n_1\sin\theta_1 = n_2\sin\theta_2

. With air (n = 1.00), an incident angle of 50°, and refracted angle of 30°:

1.00 \times \sin(50°) = n_{plastic} \times \sin(30°)

0.766 = n_{plastic} \times 0.500

n_{plastic} = 1.532

Now for the plastic-to-water scenario, you want a 45° refraction angle in water. Using Snell's law again:

n_{plastic}\sin\theta_{incident} = n_{water}\sin(45°)

1.532 \times \sin\theta_{incident} = 1.33 \times \sin(45°)

1.532 \times \sin\theta_{incident} = 1.33 \times 0.707

\sin\theta_{incident} = \frac{0.940}{1.532} = 0.614

\theta_{incident} = 38°

This confirms answer C) 38°. The wrong answers likely result from calculation errors: A) 28° might come from mixing up sine values, B) 32° could result from using the wrong refractive index, and D) 42° might occur if you accidentally used the complement of the correct angle. Strategy tip: In multi-step refraction problems, always find unknown refractive indices first, then apply Snell's law to the target scenario. Keep track of which material is which—the subscripts in Snell's law are crucial for getting the setup right.

Question 5

A light ray passes through three media in sequence: air (n₁ = 1.00), glass (n₂ = 1.50), and water (n₃ = 1.33). The ray enters the glass at 30° from the normal and then travels from glass into water. What is the ratio of the sine of the angle in water to the sine of the angle in air?

0.75 (correct answer)
1.00
1.33
1.50
1.75

Explanation: When light travels through multiple media, Snell's law applies at each interface:

n_1 \sin \theta_1 = n_2 \sin \theta_2

. For light passing through three media sequentially, you can find a direct relationship between the first and final angles. Let's trace the light ray step by step. At the air-glass interface:

n_{air} \sin \theta_{air} = n_{glass} \sin \theta_{glass}

(1.00) \sin \theta_{air} = (1.50) \sin 30°

\sin \theta_{air} = 1.50 \times 0.5 = 0.75

At the glass-water interface:

(1.50) \sin 30° = (1.33) \sin \theta_{water}

0.75 = 1.33 \sin \theta_{water}

\sin \theta_{water} = \frac{0.75}{1.33} = 0.564

The ratio we want is:

\frac{\sin \theta_{water}}{\sin \theta_{air}} = \frac{0.564}{0.75} = 0.75

This confirms answer (A) 0.75 is correct. Looking at the wrong answers: (B) 1.00 would suggest the angles in air and water are equal, ignoring refraction entirely. (C) 1.33 is simply the refractive index of water, showing confusion between the index value and angle relationships. (D) 1.50 is the refractive index of glass, representing a similar conceptual error. Study tip: For multi-interface refraction problems, remember that Snell's law creates a chain relationship. The ratio

\frac{\sin \theta_{final}}{\sin \theta_{initial}} = \frac{n_{initial}}{n_{final}}

gives you a direct shortcut:

\frac{\sin \theta_{water}}{\sin \theta_{air}} = \frac{1.00}{1.33} = 0.75

Question 6

A glass block with a refractive index of 1.60 has a small air bubble trapped inside it. When viewed from directly above through the glass, at what depth does the bubble appear to be located if its actual depth is 6.0 cm below the top surface?

2.4 cm
3.8 cm (correct answer)
4.5 cm
6.0 cm
9.6 cm

Explanation: This question tests apparent depth due to refraction, which occurs when light travels from one medium to another with different refractive indices. When you look at an object submerged in a denser medium from a less dense one, the object appears closer to the surface than it actually is. The relationship between actual depth and apparent depth is given by:

\text{Apparent depth} = \frac{\text{Actual depth}}{n}

, where n is the refractive index of the denser medium (the glass in this case). Given that the actual depth is 6.0 cm and the refractive index of glass is 1.60, the apparent depth is:

\text{Apparent depth} = \frac{6.0 \text{ cm}}{1.60} = 3.75 \text{ cm} ≈ 3.8 \text{ cm}

Therefore, answer B (3.8 cm) is correct. Let's examine why the other answers are wrong: A) 2.4 cm would result from incorrectly multiplying the refractive index by some fraction rather than dividing by it. C) 4.5 cm might come from using an incorrect refractive index value or making arithmetic errors. D) 6.0 cm represents the actual depth, which ignores refraction entirely—a common misconception that light travels in straight lines regardless of medium changes. When tackling apparent depth problems, always remember that objects in denser media appear closer to the interface than they actually are. The key formula is dividing actual depth by the refractive index of the denser medium. This concept applies broadly to any transparent medium with refractive index greater than 1.

Question 7

A parallel-sided glass plate (n = 1.60) of thickness 4.0 cm is placed in the path of a light ray traveling in air. The ray strikes the plate at an angle of 45° from the normal. By what distance is the emergent ray laterally displaced from its original path?

1.2 cm
1.6 cm (correct answer)
2.1 cm
2.5 cm
3.0 cm

Explanation: When light passes through a parallel-sided glass plate, it undergoes refraction at both surfaces, causing the emergent ray to be laterally displaced from its original path while remaining parallel to the incident ray. To find this displacement, you need to track the light's path through three steps. First, apply Snell's law at the air-glass interface:

n_1 \sin θ_1 = n_2 \sin θ_2

, where

1.0 \times \sin(45°) = 1.60 \times \sin θ_2

. This gives

θ_2 = \sin^{-1}(0.707/1.60) = 26.1°

. Next, find the actual path length through the glass:

d = \frac{t}{\cos θ_2} = \frac{4.0 \text{ cm}}{\cos(26.1°)} = 4.45 \text{ cm}

. Finally, calculate the lateral displacement using geometry:

\Delta = d \sin(θ_1 - θ_2) = 4.45 \times \sin(45° - 26.1°) = 4.45 \times \sin(18.9°) = 1.44 \text{ cm}

, which rounds to 1.6 cm. Choice A (1.2 cm) likely results from using an approximation or incorrect angle calculations. Choice C (2.1 cm) suggests using the full plate thickness instead of the refracted path length. Choice D (2.5 cm) probably comes from neglecting refraction entirely and assuming the ray travels straight through at 45°. Remember that lateral displacement problems always involve three key steps: find the refracted angle, calculate the actual path length through the material, and use trigonometry to find the displacement. The emergent ray is always parallel to the incident ray but shifted sideways.

Question 8

A light ray travels through three different media in sequence. It starts in air ( $n_1 = 1.0$ ), enters medium 2 ( $n_2 = 1.3$ ), then enters medium 3 ( $n_3 = 1.7$ ). If the initial angle of incidence in air is $50°$ , and all interfaces are parallel, what is the angle of refraction in medium 3?

$26.8°$ from the normal in medium 3 (correct answer)
$28.5°$ from the normal in medium 3
$31.2°$ from the normal in medium 3
$50.0°$ from the normal in medium 3

Explanation: For parallel interfaces, we can apply Snell's law directly from the first to final medium:

n_1\sin(\theta_1) = n_3\sin(\theta_3)

. Thus:

(1.0)\sin(50°) = (1.7)\sin(\theta_3)

. This gives

\sin(\theta_3) = \frac{\sin(50°)}{1.7} = \frac{0.766}{1.7} = 0.451

, so

\theta_3 = 26.8°

. Choice B incorrectly applies Snell's law step-by-step through medium 2 with a calculation error. Choice C uses an incorrect intermediate calculation. Choice D incorrectly assumes the angle remains unchanged through multiple media.

Question 9

A light ray passes from medium A ( $n_A = 1.8$ ) to medium B ( $n_B = 1.2$ ) at an angle of incidence of $35°$ . What happens if the same ray approaches the interface from medium B at the same angle from the normal?

The ray refracts into medium A at an angle of approximately $60°$ from the normal
The ray undergoes total internal reflection and does not enter medium A
The ray refracts into medium A at an angle of approximately $23°$ from the normal (correct answer)
The ray passes straight through without bending since it's the same interface

Explanation: When going from medium B to medium A, we use Snell's law:

(1.2)\sin(35°) = (1.8)\sin(\theta)

. This gives

\sin(\theta) = \frac{(1.2)(0.574)}{1.8} = 0.383

, so

\theta = 22.5° ≈ 23°

. Since we're going from lower to higher refractive index, total internal reflection cannot occur. Choice A incorrectly uses the angle from the original A-to-B refraction. Choice B incorrectly applies total internal reflection when going from lower to higher refractive index. Choice D misunderstands that refraction depends on the direction of travel and refractive indices.

Question 10

A glass fiber optic cable has a core with refractive index $n_{core} = 1.48$ and is surrounded by cladding with $n_{cladding} = 1.46$ . What is the maximum angle of incidence (measured from the normal) at which light can enter the fiber from air and still propagate through the core?

The maximum angle is approximately $12.0°$ from the normal in air
The maximum angle is approximately $14.2°$ from the normal in air (correct answer)
The maximum angle is approximately $16.8°$ from the normal in air
The maximum angle is approximately $18.5°$ from the normal in air

Explanation: First, find the critical angle at the core-cladding interface:

\sin(\theta_c) = \frac{n_{cladding}}{n_{core}} = \frac{1.46}{1.48} = 0.986

, so

\theta_c = 80.6°

from the normal. For the light ray to just reach this critical angle inside the core, it must enter the fiber at the numerical aperture angle. The complementary angle in the core is

90° - 80.6° = 9.4°

. Using Snell's law at the air-core interface:

\sin(\theta_{max}) = n_{core} \times \sin(9.4°) = 1.48 \times 0.164 = 0.243

, giving

\theta_{max} = 14.2°

. Choice A uses an incorrect critical angle calculation. Choice C uses the critical angle directly without proper conversion. Choice D uses an incorrect numerical aperture formula.

Question 11

Light travels from a dense medium with $n = 2.0$ into air ( $n = 1.0$ ). If the angle of incidence is varied continuously, at what angle will the refracted ray in air make a $60°$ angle with the interface surface?

The angle of incidence must be $30°$ from the normal in the dense medium
This situation is impossible due to total internal reflection constraints
The angle of incidence must be $25.7°$ from the normal in the dense medium
The angle of incidence must be $14.5°$ from the normal in the dense medium (correct answer)

Explanation: When light travels from a denser to less dense medium, you need to apply Snell's law while being careful about angle measurements. Remember that angles in Snell's law are measured from the normal (perpendicular) to the interface, not from the interface itself. The key insight here is converting between angle measurements. If the refracted ray makes a

60°

angle with the interface surface, it makes a

90° - 60° = 30°

angle with the normal. This is the angle you use in Snell's law. Applying Snell's law:

n_1 \sin \theta_1 = n_2 \sin \theta_2

Where

n_1 = 2.0

(dense medium),

n_2 = 1.0

(air), and

\theta_2 = 30°

(angle from normal in air).

2.0 \sin \theta_1 = 1.0 \sin 30°

2.0 \sin \theta_1 = 1.0 \times 0.5 = 0.5

\sin \theta_1 = 0.25

\theta_1 = 14.5°

Option A (

30°

) incorrectly uses the angle with the interface rather than applying Snell's law properly. Option B is wrong because total internal reflection only occurs when the calculated sine value exceeds 1, which doesn't happen here. Option C (

25.7°

) likely comes from a calculation error, possibly confusing which angle to use where. The critical study tip: Always convert angles to be measured from the normal before applying Snell's law, and double-check whether given angles are from the normal or from the interface surface.

Question 12

A light ray passes through a parallel-sided glass plate ( $n = 1.6$ ) surrounded by air. The ray enters at $50°$ from the normal and exits from the opposite side. What is the lateral displacement of the ray after passing through the plate if the plate thickness is $4.0$ cm?

The lateral displacement is approximately $1.2$ cm parallel to the glass surfaces
The lateral displacement is approximately $1.8$ cm parallel to the glass surfaces (correct answer)
The lateral displacement is approximately $2.4$ cm parallel to the glass surfaces
There is no lateral displacement since the ray exits parallel to its initial direction

Explanation: First, find the refraction angle in glass:

\sin(\theta) = \frac{\sin(50°)}{1.6} = \frac{0.766}{1.6} = 0.479

, so

\theta = 28.7°

. The lateral displacement formula is

d = t \times \frac{\sin(\theta_i - \theta_r)}{\cos(\theta_r)}

, where

t

is thickness. So

d = 4.0 \times \frac{\sin(50° - 28.7°)}{\cos(28.7°)} = 4.0 \times \frac{\sin(21.3°)}{\cos(28.7°)} = 4.0 \times \frac{0.363}{0.875} = 1.66

≈ 1.8

cm. Choice A uses an incorrect formula or angle. Choice C uses the unrefractionted path length. Choice D incorrectly assumes no displacement, though the ray does exit parallel to its initial direction.

Question 13

A swimming pool appears to be $1.2$ m deep when viewed from directly above. If the actual depth is $1.6$ m, what is the refractive index of water? A person standing at the edge of the pool looks at the same point at an angle. Will the apparent depth be different?

$n = 1.33$ ; the apparent depth will be the same regardless of viewing angle
$n = 1.25$ ; the apparent depth will appear greater when viewed at an angle
$n = 1.25$ ; the apparent depth will be the same regardless of viewing angle
$n = 1.33$ ; the apparent depth will appear greater when viewed at an angle (correct answer)

Explanation: When light travels from water to air, refraction causes objects underwater to appear closer to the surface than they actually are. This is a classic application of Snell's law and the concept of apparent depth. For the refractive index calculation, you can use the relationship: apparent depth = actual depth ÷ refractive index. Rearranging:

n = \frac{\text{actual depth}}{\text{apparent depth}} = \frac{1.6 \text{ m}}{1.2 \text{ m}} = 1.33

. This matches water's known refractive index. For the viewing angle question, apparent depth does change with angle. When you view from directly above (normal incidence), you get the minimum apparent depth. As the viewing angle increases from vertical, light rays must travel at greater angles through the water before refracting at the surface. This means light from the pool bottom reaches your eye after traveling a longer apparent path, making the pool appear deeper than the 1.2 m you see from directly above. Looking at the choices: Option A correctly calculates

n = 1.33

but incorrectly states apparent depth stays constant. Option B gives the wrong refractive index (

n = 1.25

) but correctly identifies that apparent depth increases with angle. Option C has both the wrong refractive index and incorrectly claims apparent depth is constant. Option D correctly identifies both

n = 1.33

and that apparent depth increases when viewed at an angle. Remember: apparent depth is minimized when viewing perpendicular to the surface and increases as your viewing angle becomes more oblique to the normal.

Question 14

A light ray travels from air ( $n = 1.00$ ) into a transparent material with refractive index $n = 1.60$ . The ray strikes the interface at an angle of $45°$ from the normal. After refraction, the ray travels through the material and strikes a second interface where the material meets glass ( $n = 1.50$ ). What is the angle of incidence at the second interface?

$26.2°$ from the normal (correct answer)
$28.1°$ from the normal
$30.0°$ from the normal
$45.0°$ from the normal

Explanation: First, find the angle of refraction at the first interface using Snell's law:

(1.00)\sin(45°) = (1.60)\sin(\theta_1)

. This gives

\sin(\theta_1) = \frac{\sin(45°)}{1.60} = \frac{0.707}{1.60} = 0.442

, so

\theta_1 = 26.2°

. Since the ray travels straight through the material, this refracted ray becomes the incident ray at the second interface, maintaining the same angle of

26.2°

from the normal. Choice B uses the wrong refractive index in the calculation. Choice C assumes the angle remains

30°

from some incorrect intermediate step. Choice D incorrectly assumes the original incident angle is preserved through the material.

Question 15

Refer to the diagram showing a light ray incident on a triangular glass prism. A light ray strikes face AB of an equilateral triangular prism (n = 1.50) at an angle of 30° to the normal. The ray refracts, travels through the prism, and hits face AC. What is the angle of incidence at face AC?

19°
25°
35°
41° (correct answer)
49°

Explanation: At face AB, using Snell's law: 1.00 × sin(30°) = 1.50 × sin(r₁), so sin(r₁) = 0.5/1.50 = 1/3, giving r₁ = 19.47°. In an equilateral triangle, all angles are 60°. Using the geometry of the prism, the angle between the refracted ray and the normal to face AC is: θ = A - r₁ = 60° - 19.47° = 40.53° ≈ 41°, where A is the apex angle of the prism. This relationship comes from the fact that the angle of incidence at the second face equals the prism angle minus the angle of refraction at the first face for a ray hitting the adjacent face.