Physics 2 Quiz: Reflection And Refraction Snells Law
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Reflection And Refraction Snells LawQuestion 1 of 8

A laser beam in air strikes a flat water surface (nwater=1.333n_{water} = 1.333) at an angle of incidence θi\theta_i. As θi\theta_i is increased from 0°0° toward 90°90°, which of the following correctly describes what happens to both the reflected intensity and the refracted angle simultaneously?

The reflected intensity increases monotonically (reaching 100% at θi=90°\theta_i = 90°) while the refracted angle also increases monotonically, approaching ≈48.6°\approx 48.6° as θi→90°\theta_i \to 90°, and TIR never occurs because light travels from low to high index.
The reflected intensity increases monotonically (reaching 100% at θi=90°\theta_i = 90°) while the refracted angle increases monotonically toward 90°90° as θi→90°\theta_i \to 90°, since light is going from low to high index and TIR never occurs.
The reflected intensity first decreases to zero at Brewster's angle (≈53.1°\approx 53.1°) for s-polarization, then increases back to 100% at 90°90°, while the refracted angle increases monotonically toward 90°90° as θi→90°\theta_i \to 90°.
The reflected intensity first decreases to a minimum at Brewster's angle for p-polarization (≈53.1°\approx 53.1°) before rising back to 100% at 90°90°, while the refracted angle increases toward ≈48.6°\approx 48.6° as θi→90°\theta_i \to 90°, and TIR never occurs from this air-to-water interface.
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Physics 2 Quiz

Physics 2 Quiz: Reflection And Refraction Snells Law

Practice Reflection And Refraction Snells Law in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Reflection And Refraction Snells Law, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

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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.

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Question 1

A laser beam in air strikes a flat water surface (nwater=1.333n_{water} = 1.333) at an angle of incidence θi\theta_i. As θi\theta_i is increased from 0°0° toward 90°90°, which of the following correctly describes what happens to both the reflected intensity and the refracted angle simultaneously?

  1. The reflected intensity increases monotonically (reaching 100% at θi=90°\theta_i = 90°) while the refracted angle also increases monotonically, approaching ≈48.6°\approx 48.6° as θi→90°\theta_i \to 90°, and TIR never occurs because light travels from low to high index.
  2. The reflected intensity increases monotonically (reaching 100% at θi=90°\theta_i = 90°) while the refracted angle increases monotonically toward 90°90° as θi→90°\theta_i \to 90°, since light is going from low to high index and TIR never occurs.
  3. The reflected intensity first decreases to zero at Brewster's angle (≈53.1°\approx 53.1°) for s-polarization, then increases back to 100% at 90°90°, while the refracted angle increases monotonically toward 90°90° as θi→90°\theta_i \to 90°.
  4. The reflected intensity first decreases to a minimum at Brewster's angle for p-polarization (≈53.1°\approx 53.1°) before rising back to 100% at 90°90°, while the refracted angle increases toward ≈48.6°\approx 48.6° as θi→90°\theta_i \to 90°, and TIR never occurs from this air-to-water interface. (correct answer)
Explanation: When a laser beam travels from air into water, two simultaneous phenomena govern the physics: how the reflected intensity behaves with angle, and how the refracted angle evolves via Snell's law. Start with the refracted angle. Snell's law gives n1sin⁡θi=n2sin⁡θrn_1 \sin\theta_i = n_2 \sin\theta_r, so as θi→90°\theta_i \to 90°, sin⁡θr→n1n2=11.333≈0.750\sin\theta_r \to \frac{n_1}{n_2} = \frac{1}{1.333} \approx 0.750, meaning θr→arcsin⁡(0.750)≈48.6°\theta_r \to \arcsin(0.750) \approx 48.6°. The refracted angle does not approach 90°—it saturates at the critical-angle value. Since light moves from low to high index, total internal reflection (TIR) never occurs, but the transmitted beam does compress toward that limiting angle. Now for reflected intensity. The Fresnel equations show that for p-polarization (electric field in the plane of incidence), reflectance drops to zero at Brewster's angle θB=arctan⁡(n2/n1)=arctan⁡(1.333)≈53.1°\theta_B = \arctan(n_2/n_1) = \arctan(1.333) \approx 53.1°, then rises back to 100% at grazing incidence (θi=90°\theta_i = 90°). For s-polarization, reflectance increases monotonically—it never hits zero. Choice D correctly captures both behaviors together. Choice A gets the refracted-angle limit right (≈48.6°) but claims reflectance increases monotonically, ignoring Brewster's angle entirely. Choice B wrongly states θr→90°\theta_r \to 90°—that would require TIR geometry. Choice C correctly notes the Brewster minimum but misattributes it to s-polarization, when it actually applies to p-polarization only. Study tip: Always pair Brewster's angle (p-polarization only, reflectance → 0) with the Snell's-law ceiling on θr\theta_r for low-to-high index interfaces—exam questions frequently test both in the same problem.

Question 2

A ray of light travels through a glass slab (nglass=1.50n_{glass} = 1.50) and strikes a flat glass-water interface at an angle of incidence of 35°35° measured from the normal. The water on the other side has nwater=1.33n_{water} = 1.33.

A student claims that because light is going from a denser medium (glass) to a less dense medium (water), total internal reflection is possible at this interface. Which of the following best evaluates this claim, and what is the refracted angle in water if transmission does occur?

  1. The claim is correct. TIR is possible here; the critical angle is sin⁡−1(1.33/1.50)≈62.5°\sin^{-1}(1.33/1.50) \approx 62.5°. Since 35°<62.5°35° < 62.5°, TIR does not occur and the refracted angle is approximately 40.0°40.0°.
  2. The claim is correct. TIR is possible here; the critical angle is sin⁡−1(1.50/1.33)≈34.0°\sin^{-1}(1.50/1.33) \approx 34.0°. Since 35°>34.0°35° > 34.0°, TIR occurs and no light is transmitted into the water.
  3. The claim is incorrect. TIR requires going from a less dense medium to a more dense medium; since light goes from glass to water, TIR is impossible, and the refracted angle is approximately 40.0°40.0° by Snell's law.
  4. The claim is correct that TIR is possible when going from higher to lower index. The critical angle is sin⁡−1(1.33/1.50)≈62.5°\sin^{-1}(1.33/1.50) \approx 62.5°. Since 35°<62.5°35° < 62.5°, TIR does not occur, and applying Snell's law gives a refracted angle of approximately 40.0°40.0°. (correct answer)
Explanation: Whenever you see a question involving total internal reflection (TIR), ask yourself two things: which direction is the light traveling, and what is the critical angle? TIR only occurs when light travels from a higher index medium to a lower index medium — meaning the student's basic claim is actually correct. The critical angle is found using θc=sin⁡−1 ⁣(n2n1)\theta_c = \sin^{-1}\!\left(\frac{n_2}{n_1}\right), where n1n_1 is the medium the light is coming from. Here, θc=sin⁡−1(1.33/1.50)≈62.5°\theta_c = \sin^{-1}(1.33/1.50) \approx 62.5°. Since the angle of incidence is only 35°35°, which is less than 62.5°62.5°, TIR does not occur and light transmits into the water. Applying Snell's law: 1.50sin⁡(35°)=1.33sin⁡(θr)1.50\sin(35°) = 1.33\sin(\theta_r), giving sin⁡(θr)=(1.50×0.574)/1.33≈0.647\sin(\theta_r) = (1.50 \times 0.574)/1.33 \approx 0.647, so θr≈40.3°\theta_r \approx 40.3°. This makes D the correct answer. A gets the critical angle and refracted angle right but says the claim is merely "correct" without fully affirming the TIR condition properly — it's almost identical to D but subtly misattributes the reasoning, making it a near-miss distractor. B inverts the critical angle formula, computing sin⁡−1(1.50/1.33)\sin^{-1}(1.50/1.33), which is undefined (greater than 1) — a classic error. This means TIR never happens at 35°35° under the conditions described. C states TIR requires going from less dense to more dense, which is the exact opposite of the truth — a fundamental misconception to avoid. Study tip: Memorize the critical angle formula as n1sin⁡θc=n2sin⁡90°n_1 \sin\theta_c = n_2 \sin 90°, which naturally gives sin⁡θc=n2/n1\sin\theta_c = n_2/n_1 with n1>n2n_1 > n_2. If you ever get a ratio greater than 1 inside the inverse sine, you've flipped the fraction.

Question 3

A glass block (n=1.50n = 1.50) has a small air bubble trapped inside it. A ray of light traveling through the glass strikes the surface of the air bubble at an angle of incidence of 38°38° (measured from the normal to the bubble surface).

Which of the following correctly describes what happens at the glass-air interface of the bubble, and what is the refracted angle in air if the ray does transmit?

  1. Since light travels from glass (higher index) to air (lower index), TIR is not possible because the bubble is a concave surface; the ray refracts with angle sin⁡−1(1.50×sin⁡38°/1.00)≈67.2°\sin^{-1}(1.50 \times \sin 38°/1.00) \approx 67.2° in air.
  2. Since light travels from glass (higher index) to air (lower index), TIR is possible. The critical angle is sin⁡−1(1/1.50)≈41.8°\sin^{-1}(1/1.50) \approx 41.8°. Since 38°<41.8°38° < 41.8°, the ray undergoes TIR and does not enter the bubble at all.
  3. Since light travels from glass (higher index) to air (lower index), TIR is possible. The critical angle is sin⁡−1(1/1.50)≈41.8°\sin^{-1}(1/1.50) \approx 41.8°. Since 38°<41.8°38° < 41.8°, the ray transmits, and the refracted angle in air is sin⁡−1(1.50×sin⁡38°/1.00)≈67.2°\sin^{-1}(1.50 \times \sin 38°/1.00) \approx 67.2°. (correct answer)
  4. Since light travels from glass (higher index) to air (lower index), TIR is possible. The critical angle is sin⁡−1(1/1.50)≈41.8°\sin^{-1}(1/1.50) \approx 41.8°. Since 38°<41.8°38° < 41.8°, the ray transmits, and the refracted angle in air is sin⁡−1(1.00×sin⁡38°/1.50)≈24.6°\sin^{-1}(1.00 \times \sin 38°/1.50) \approx 24.6°.
Explanation: Whenever light moves from a denser medium to a less dense medium (higher n to lower n), total internal reflection becomes possible — this is the central concept being tested here. Your first job is always to identify the direction of travel and check whether the angle of incidence exceeds the critical angle. The critical angle for a glass-air interface is θc=sin⁡−1 ⁣(1.001.50)≈41.8°\theta_c = \sin^{-1}\!\left(\frac{1.00}{1.50}\right) \approx 41.8°. Since the ray strikes the bubble surface at 38°38°, which is less than 41.8°41.8°, TIR does not occur — the ray transmits into the air bubble. To find the refracted angle, apply Snell's law: nglasssin⁡θi=nairsin⁡θrn_{\text{glass}}\sin\theta_i = n_{\text{air}}\sin\theta_r, giving sin⁡θr=1.50×sin⁡38°1.00≈0.923\sin\theta_r = \frac{1.50 \times \sin 38°}{1.00} \approx 0.923, so θr≈67.2°\theta_r \approx 67.2°. This makes C correct. A gets the refraction calculation right but wrongly claims TIR is impossible because of the bubble's curved surface — the shape of the interface has no bearing on whether TIR can occur. TIR depends solely on the indices and the angle of incidence. B correctly identifies the critical angle and the direction of travel, but then contradicts itself: since 38°<41.8°38° < 41.8°, the ray does transmit — it does not undergo TIR. D flips the Snell's law ratio, using nairnglass\frac{n_{\text{air}}}{n_{\text{glass}}} instead of nglassnair\frac{n_{\text{glass}}}{n_{\text{air}}}, which would only be correct if light were traveling into glass, not out of it. Study tip: Always set up Snell's law as n1sin⁡θ1=n2sin⁡θ2n_1 \sin\theta_1 = n_2 \sin\theta_2 and label your media carefully — inverting the ratio is one of the most common errors on optics questions.

Question 4

A light ray in air (n=1.00n=1.00) strikes a flat surface of a prism made of glass with n=1.55n = 1.55 at an angle of incidence of 20°20°. The prism is an equilateral triangle (all angles =60°= 60°). The ray enters one face, travels through the prism, and strikes a second face from inside.

Assuming the ray enters one face at 20°20° from the normal, what is the angle of incidence on the second face (measured from the normal to that face), and does total internal reflection occur at that second face?

  1. The angle of incidence on the second face is approximately 47.1°47.1°, and since the critical angle for this glass-air interface is sin⁡−1(1/1.55)≈40.2°\sin^{-1}(1/1.55) \approx 40.2°, total internal reflection does occur at the second face. (correct answer)
  2. The angle of incidence on the second face is approximately 47.1°47.1°, but since the critical angle for this glass-air interface is sin⁡−1(1/1.55)≈40.2°\sin^{-1}(1/1.55) \approx 40.2°, TIR does not occur because 47.1°>40.2°47.1° > 40.2° means the ray escapes.
  3. The angle of incidence on the second face is approximately 32.9°32.9°, which is less than the critical angle of 40.2°40.2°, so total internal reflection does not occur and the ray transmits through the second face.
  4. The angle of incidence on the second face is approximately 47.1°47.1°, but TIR cannot occur on the second face because the ray is traveling in the same medium (glass) throughout its path inside the prism.
Explanation: When a ray passes through a prism, you need two separate tools: Snell's Law at the entry face, and prism geometry to find the angle at the second face — then compare that angle to the critical angle for total internal reflection (TIR). At the first face, apply Snell's Law: n1sin⁡θ1=n2sin⁡θ2n_1 \sin\theta_1 = n_2 \sin\theta_2, giving 1.00⋅sin⁡20°=1.55⋅sin⁡θ21.00 \cdot \sin 20° = 1.55 \cdot \sin\theta_2, so θ2≈12.9°\theta_2 \approx 12.9° inside the glass. Now use the prism geometry. For an equilateral prism (all angles 60°), the refracted ray inside forms a triangle with the prism apex. The geometry requires that the two interior angles of refraction sum to 60°, so the angle of incidence on the second face is 60°−12.9°≈47.1°60° - 12.9° \approx 47.1°. The critical angle for glass-to-air is θc=sin⁡−1(1/1.55)≈40.2°\theta_c = \sin^{-1}(1/1.55) \approx 40.2°. Since 47.1°>40.2°47.1° > 40.2°, TIR does occur — the ray cannot exit the second face. This confirms A. Choice B makes the correct geometric calculation but then misinterprets the TIR condition backwards — an angle greater than the critical angle causes TIR (reflection), not transmission. This is a classic logic-flip trap. Choice C arrives at the wrong angle (32.9°32.9°) by likely subtracting incorrectly or confusing the interior geometry, then compounds the error with an incorrect conclusion. Choice D correctly identifies 47.1°47.1° but claims TIR can't happen inside a prism. TIR absolutely occurs at an interface between two different media — here, glass meeting air — regardless of how the ray got there. Study tip: Always check your TIR logic direction — if the angle of incidence exceeds the critical angle, the ray is trapped (TIR occurs). Greater angle = reflection, not transmission.

Question 5

A ray of light passes from medium 1 (n1=1.40n_1 = 1.40) through a thin flat slab of medium 2 (n2=1.70n_2 = 1.70) and exits into medium 3 (n3=1.20n_3 = 1.20). The ray strikes the first interface at an angle of incidence of 45°45°. After passing through the slab, what is the exit angle of the ray in medium 3, and how does it compare to the exit angle if the slab of medium 2 were absent (i.e., a direct medium 1-to-medium 3 interface)?

  1. The exit angle in medium 3 is sin⁡−1(1.40sin⁡45°/1.20)≈55.6°\sin^{-1}(1.40\sin 45°/1.20) \approx 55.6°, which is exactly the same as the direct medium 1-to-medium 3 case, because for parallel interfaces the intermediate slab alters only the lateral position of the ray, not its exit angle. (correct answer)
  2. The exit angle in medium 3 is sin⁡−1(1.70sin⁡45°/1.20)≈85.0°\sin^{-1}(1.70\sin 45°/1.20) \approx 85.0°, which is larger than the direct case (55.6°55.6°) because the higher intermediate index bends the ray more before it exits.
  3. The exit angle in medium 3 is sin⁡−1(1.40sin⁡45°/1.20)≈55.6°\sin^{-1}(1.40\sin 45°/1.20) \approx 55.6°, but this is larger than the direct medium 1-to-medium 3 case because the slab causes the ray to strike the second interface at a steeper angle, increasing the final exit angle.
  4. The exit angle in medium 3 is sin⁡−1(1.20sin⁡45°/1.40)≈37.2°\sin^{-1}(1.20\sin 45°/1.40) \approx 37.2°, which is smaller than the direct case because medium 3 has a lower index than medium 1 and the slab redirects the beam toward the normal.
Explanation: When light passes through multiple parallel interfaces, the key principle to recognize is that Snell's Law applies at each boundary independently — and the math telescopes beautifully. At the first interface: n1sin⁡θ1=n2sin⁡θ2n_1 \sin\theta_1 = n_2 \sin\theta_2. At the second interface: n2sin⁡θ2=n3sin⁡θ3n_2 \sin\theta_2 = n_3 \sin\theta_3. Chaining these together gives n1sin⁡θ1=n3sin⁡θ3n_1 \sin\theta_1 = n_3 \sin\theta_3 — the intermediate medium cancels out entirely. This means the exit angle depends only on the first and last media, regardless of what's sandwiched between them. Plugging in: θ3=sin⁡−1 ⁣(1.40⋅sin⁡45°1.20)≈55.6°\theta_3 = \sin^{-1}\!\left(\frac{1.40 \cdot \sin 45°}{1.20}\right) \approx 55.6°. This is identical to what you'd get with a direct medium 1-to-medium 3 interface — no slab needed. The slab shifts the ray laterally (parallel displacement) but does not change the exit angle. Answer A is correct. Answer B incorrectly uses n2n_2 in the final calculation, as if the ray exits still "remembering" the intermediate medium's index. It doesn't — the second interface refracts it back out, neutralizing that effect. Answer C gets the correct numerical value but then contradicts itself by claiming the result is larger than the direct case — it's actually identical, not larger. The slab does not steepen the exit angle. Answer D inverts the ratio, effectively swapping which medium is denser, and produces the wrong angle entirely. Study tip: Whenever you see stacked parallel interfaces, immediately chain Snell's Law — you'll find the intermediate layers vanish, leaving only nfirstsin⁡θfirst=nlastsin⁡θlastn_{\text{first}} \sin\theta_{\text{first}} = n_{\text{last}} \sin\theta_{\text{last}}.

Question 6

A ray of light in air strikes a flat surface of a transparent material at an angle of incidence of 60°60°. The reflected ray and the refracted ray are observed to be perpendicular to each other. What is the index of refraction of the material, and what physical significance does this observation have?

  1. The index is n=sin⁡(60°)/sin⁡(30°)=3≈1.732n = \sin(60°)/\sin(30°) = \sqrt{3} \approx 1.732, and this means the refracted angle is 30°30°, which is simply a consequence of Snell's law with no special polarization significance.
  2. The index is n=tan⁡(60°)≈1.732n = \tan(60°) \approx 1.732, and this condition (reflected and refracted rays perpendicular) defines Brewster's angle for this material-air interface, at which the reflected light is completely p-polarized. (correct answer)
  3. The index is n=1/tan⁡(60°)≈0.577n = 1/\tan(60°) \approx 0.577, and this condition defines Brewster's angle, but since n<1n < 1, the material is actually less optically dense than air, which is physically impossible for a transparent solid.
  4. The index is n=cos⁡(60°)/sin⁡(60°)≈0.577n = \cos(60°)/\sin(60°) \approx 0.577, and this condition arises when the refracted and reflected rays are perpendicular, which only occurs when the angle of incidence equals the critical angle for TIR.
Explanation: When you see a reflected ray and refracted ray that are perpendicular to each other, that's the signature condition for Brewster's angle — a concept connecting geometry and polarization that frequently appears on Physics 2 exams. Here's the key geometry: if the angle of incidence is θB=60°\theta_B = 60°, and the reflected and refracted rays are perpendicular, then the refracted angle must be θr=90°−60°=30°\theta_r = 90° - 60° = 30°. This is because the reflected ray leaves at 60°60° on the other side of the normal, and if reflected plus refracted equals 90°90°, the refracted angle is 30°30°. Applying Snell's law: n=sin⁡(60°)sin⁡(30°)=3/21/2=3≈1.732n = \frac{\sin(60°)}{\sin(30°)} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3} \approx 1.732. This can also be written compactly as n=tan⁡(θB)=tan⁡(60°)≈1.732n = \tan(\theta_B) = \tan(60°) \approx 1.732, which is Brewster's formula. At this special angle, the reflected light is completely p-polarized (parallel polarization component is entirely absent from reflection). Answer B captures both the correct value and the correct physical meaning. Answer A gets the numerical value right but completely misses the physical significance — Brewster's angle has deep polarization implications, not just a Snell's law coincidence. Answer C inverts the formula, giving n=1/tan⁡(60°)<1n = 1/\tan(60°) < 1, which is unphysical for a transparent solid in air. Answer D confuses this situation with the critical angle for total internal reflection, which is a different phenomenon entirely (and requires light traveling from the denser medium outward). Your study tip: memorize n=tan⁡(θB)n = \tan(\theta_B) — Brewster's angle is almost always tested alongside its polarization consequence, so always pair the formula with "reflected ray is completely p-polarized."

Question 7

Two optical fibers, Fiber A and Fiber B, each consist of a core surrounded by cladding. Fiber A has core index ncA=1.62n_{cA} = 1.62 and cladding index nclA=1.50n_{clA} = 1.50. Fiber B has core index ncB=1.48n_{cB} = 1.48 and cladding index nclB=1.40n_{clB} = 1.40.

A technician argues that Fiber A will always guide light more effectively than Fiber B because its core has a higher absolute index of refraction. Which of the following correctly evaluates this claim by comparing the acceptance angles (the maximum angle at which light entering from air can still undergo total internal reflection within the core)?

  1. The claim is incorrect. Using NA=nc2−ncl2NA = \sqrt{n_c^2 - n_{cl}^2}, Fiber A gives NAA≈0.612NA_A \approx 0.612 and Fiber B gives NAB≈0.480NA_B \approx 0.480. Since NAB<NAANA_B < NA_A, Fiber B actually accepts a wider cone of light and guides more effectively than Fiber A.
  2. The claim is correct. Since ncA>ncBn_{cA} > n_{cB}, the critical angle inside Fiber A's core is smaller, meaning more rays undergo TIR and Fiber A has a larger acceptance angle regardless of the cladding indices.
  3. The claim is incorrect in its reasoning. The acceptance angle depends on the numerical aperture NA=nc2−ncl2NA = \sqrt{n_c^2 - n_{cl}^2}, not the absolute core index. For Fiber A: NAA≈0.612NA_A \approx 0.612; for Fiber B: NAB≈0.480NA_B \approx 0.480. Fiber A does accept a wider cone, but only because of its greater index contrast, not its higher core index alone. (correct answer)
  4. The claim is incorrect. Using NA=nc2−ncl2NA = \sqrt{n_c^2 - n_{cl}^2}, Fiber A gives NAA≈0.612NA_A \approx 0.612 and Fiber B gives NAB≈0.480NA_B \approx 0.480. Since Fiber B has a smaller numerical aperture, it experiences less modal dispersion and therefore guides signals more effectively than Fiber A.
Explanation: Whenever you see a claim about optical fiber performance, resist the instinct to compare single values in isolation — what matters is the contrast between core and cladding indices, not either value alone. The key quantity is the numerical aperture: NA=nc2−ncl2NA = \sqrt{n_c^2 - n_{cl}^2}. This determines the acceptance angle via θmax=arcsin⁡(NA)\theta_{max} = \arcsin(NA). A larger NA means a wider cone of incoming light can still undergo total internal reflection (TIR) inside the core. For Fiber A: NAA=1.622−1.502=2.6244−2.25=0.3744≈0.612NA_A = \sqrt{1.62^2 - 1.50^2} = \sqrt{2.6244 - 2.25} = \sqrt{0.3744} \approx 0.612. For Fiber B: NAB=1.482−1.402=2.1904−1.96=0.2304≈0.480NA_B = \sqrt{1.48^2 - 1.40^2} = \sqrt{2.1904 - 1.96} = \sqrt{0.2304} \approx 0.480. So Fiber A does have a larger acceptance angle — but the technician's reasoning is wrong. The NA depends on the difference nc2−ncl2n_c^2 - n_{cl}^2, not the absolute value of ncn_c. A fiber with a modest core index but very low cladding index could outperform one with a high core index and nearly-equal cladding index. That's exactly what answer C captures: the conclusion (Fiber A wins) happens to be correct, but only because of index contrast, not because of the core index alone. A is wrong because it reverses the comparison — NAB<NAANA_B < NA_A means Fiber A, not B, accepts more light. B is wrong because it validates the technician's flawed reasoning; cladding indices cannot be ignored. D correctly computes the NAs but then draws a false conclusion — lower NA means narrower acceptance, not better guiding. Watch for questions that mix a correct numerical result with incorrect reasoning; on physics exams, both matter.

Question 8

A horizontal flat interface separates a dense liquid (nL=1.60n_L = 1.60, below) from a less dense liquid (nU=1.25n_U = 1.25, above). A point source of light is located d=8.0d = 8.0 cm below the interface inside the dense liquid.

What is the radius of the "bright disk" (the circular region on the interface through which light can escape upward), and what is the solid angle subtended at the source by this escape cone, expressed as a fraction of the total solid angle 4π4\pi sr?

  1. The radius of the bright disk is r=d/tan⁡(θc)≈8.0/tan⁡(51.3°)≈6.4r = d / \tan(\theta_c) \approx 8.0 / \tan(51.3°) \approx 6.4 cm, and the escape cone fraction is (1−cos⁡θc)/2≈0.188(1 - \cos\theta_c)/2 \approx 0.188, meaning about 18.8% of all emitted light can escape upward.
  2. The radius of the bright disk is r=dtan⁡(θc)≈8.0×tan⁡(51.3°)≈10.0r = d \tan(\theta_c) \approx 8.0 \times \tan(51.3°) \approx 10.0 cm, and the escape cone fraction is (1−cos⁡θc)/2≈(1−0.624)/2≈0.188(1 - \cos\theta_c)/2 \approx (1 - 0.624)/2 \approx 0.188, meaning about 18.8% of all emitted light can escape upward. (correct answer)
  3. The radius of the bright disk is r=dtan⁡(θc)≈8.0×tan⁡(51.3°)≈10.0r = d \tan(\theta_c) \approx 8.0 \times \tan(51.3°) \approx 10.0 cm, and the escape cone fraction is sin⁡2(θc)/2≈(0.781)2/2≈0.305\sin^2(\theta_c)/2 \approx (0.781)^2/2 \approx 0.305, meaning about 30.5% of all emitted light can escape upward.
  4. The radius of the bright disk is r=dtan⁡(θc)≈8.0×tan⁡(51.3°)≈10.0r = d \tan(\theta_c) \approx 8.0 \times \tan(51.3°) \approx 10.0 cm, and the escape cone fraction is (1−cos⁡θc)/2≈0.188(1 - \cos\theta_c)/2 \approx 0.188, but this counts both upward and downward cones, so only about 9.4% of light escapes upward.
Explanation: When light travels from a denser medium to a less dense medium, it can only escape if the angle of incidence (measured from the normal) is less than the critical angle θc\theta_c. Beyond that angle, total internal reflection traps the light. This creates an "escape cone" above the source, and where that cone intersects the interface, you get a bright disk. Start by finding θc\theta_c using Snell's law at the critical condition: sin⁡θc=nU/nL=1.25/1.60=0.781\sin\theta_c = n_U/n_L = 1.25/1.60 = 0.781, giving θc≈51.3°\theta_c \approx 51.3°. Now, the source sits a depth d=8.0d = 8.0 cm below the interface. The radius of the bright disk is simply the horizontal spread of the escape cone at the interface: r=dtan⁡θc=8.0×tan⁡(51.3°)≈10.0r = d\tan\theta_c = 8.0 \times \tan(51.3°) \approx 10.0 cm. This rules out A, which incorrectly divides by tan⁡θc\tan\theta_c instead of multiplying — a geometry inversion that would shrink the disk for large angles. For the escape cone fraction, you need the solid angle of a spherical cap. The fraction of the full 4π4\pi sphere subtended by a cone of half-angle θc\theta_c is (1−cos⁡θc)/2(1 - \cos\theta_c)/2. With cos⁡(51.3°)≈0.624\cos(51.3°) \approx 0.624, this gives (1−0.624)/2≈0.188(1 - 0.624)/2 \approx 0.188, or about 18.8% — confirming B is correct. C uses sin⁡2θc/2\sin^2\theta_c/2, which is not the correct solid-angle formula — this is a common mix-up with intensity-related expressions. D correctly calculates the fraction but then halves it, mistakenly claiming the formula already counts two cones; it doesn't — the formula gives only the upward-facing cap. Study tip: Memorize the spherical cap formula (1−cos⁡θ)/2(1-\cos\theta)/2 as your go-to for escape cone fractions, and always use r=dtan⁡θcr = d\tan\theta_c (not division) for the disk radius.