Physics 2 Quiz: Polarization

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PolarizationQuestion 1 of 10

A pile-of-plates polarizer consists of $N$ glass plates (refractive index $n = 1.50$ ) tilted at Brewster's angle for glass, so that the reflected beams are completely polarized. Each plate transmits a fraction $T_p = 1$ of the p-polarization (parallel to plane of incidence) and a fraction $T_s \approx 0.85$ of the s-polarization (perpendicular to plane of incidence) per surface at Brewster's angle.

Unpolarized light enters the pile-of-plates polarizer. After passing through $N$ plates (each with 2 surfaces), what is the degree of polarization $P = \dfrac{I_p - I_s}{I_p + I_s}$ in terms of $N$ , and how many plates are needed to achieve $P \geq 0.95$ given $T_s = 0.85$ per surface?

P = \dfrac{1 - (0.85)^{N}}{1 + (0.85)^{N}}

; approximately

N = 22

plates are needed to achieve

P \geq 0.95

P = \dfrac{1 - (0.85)^{2N}}{1 + (0.85)^{2N}}

; approximately

N = 12

plates are needed to achieve

P \geq 0.95

P = \dfrac{1 - (0.85)^{2N}}{1 + (0.85)^{2N}}

; approximately

N = 4

plates are needed to achieve

P \geq 0.95

, since each plate provides two surfaces and the degree of polarization increases rapidly.

P = 1 - (0.85)^{2N}

; approximately

N = 12

plates are needed to achieve

P \geq 0.95

, since the degree of polarization equals the fraction of s-polarization removed by the stack.

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Physics 2 Quiz

Physics 2 Quiz: Polarization

Practice Polarization in Physics 2 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 Polarization, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

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

$P = \dfrac{1 - (0.85)^{N}}{1 + (0.85)^{N}}$ ; approximately $N = 22$ plates are needed to achieve $P \geq 0.95$ .
$P = \dfrac{1 - (0.85)^{2N}}{1 + (0.85)^{2N}}$ ; approximately $N = 12$ plates are needed to achieve $P \geq 0.95$ . (correct answer)
$P = \dfrac{1 - (0.85)^{2N}}{1 + (0.85)^{2N}}$ ; approximately $N = 4$ plates are needed to achieve $P \geq 0.95$ , since each plate provides two surfaces and the degree of polarization increases rapidly.
$P = 1 - (0.85)^{2N}$ ; approximately $N = 12$ plates are needed to achieve $P \geq 0.95$ , since the degree of polarization equals the fraction of s-polarization removed by the stack.

Explanation: When unpolarized light passes through a pile-of-plates polarizer, you need to track what happens to each polarization component separately. Start with equal intensities:

I_p^0 = I_s^0 = I_0/2

. Since each plate has two surfaces, and

T_p = 1

per surface while

T_s = 0.85

per surface, after

N

plates the intensities become:

I_p = \frac{I_0}{2}(1)^{2N} = \frac{I_0}{2}, \quad I_s = \frac{I_0}{2}(0.85)^{2N}

Plugging into the degree of polarization formula:

P = \frac{I_p - I_s}{I_p + I_s} = \frac{1 - (0.85)^{2N}}{1 + (0.85)^{2N}}

To find when

P \geq 0.95

, set

\frac{1-(0.85)^{2N}}{1+(0.85)^{2N}} = 0.95

, which gives

(0.85)^{2N} \approx 0.026

. Taking logarithms:

2N \ln(0.85) = \ln(0.026)

, so

N \approx 11.6

, meaning $N = 12$ plates are needed. This confirms B is correct. A uses

(0.85)^N

instead of

(0.85)^{2N}

, forgetting that each plate contributes two surfaces — a critical geometric error that dramatically underestimates polarization efficiency, requiring 22 plates instead of 12. C gets the formula right but wildly underestimates the plate count (

N = 4

). You can check: at

N = 4

(0.85)^8 \approx 0.27

, giving

P \approx 0.57

— far below 0.95. D uses

P = 1 - (0.85)^{2N}

, which has no physical basis as a polarization formula. Degree of polarization is always a ratio of intensity differences, not a simple subtraction. Study tip: Always count surfaces, not plates — in two-surface optical elements, your exponent doubles. Write out "surfaces = 2N" explicitly before solving to avoid choice A's trap.

Question 2

A beam of unpolarized light of intensity $I_0$ passes through three ideal linear polarizers in sequence. The first polarizer has its transmission axis oriented at $0°$ from vertical. The second is oriented at $30°$ from vertical. The third is oriented at $90°$ from vertical.

What is the final intensity of the light emerging from the third polarizer, expressed as a fraction of $I_0$ ?

$\dfrac{3}{32} I_0$ , because Malus's law applies at each successive polarizer after the first, and the angles between consecutive polarizers are $30°$ and $60°$ respectively. (correct answer)
$\dfrac{3}{16} I_0$ , because the total angular span is $90°$ and one must apply $\cos^2(90°)$ to half the initial intensity, then multiply by $\cos^2(30°)$ .
$0$ , because the first and third polarizers are crossed at $90°$ , so no light can emerge regardless of any intermediate polarizer orientation.
$\dfrac{1}{8} I_0$ , because after the first polarizer the intensity is $\frac{I_0}{2}$ , and applying $\cos^2(90°)$ across the full span gives zero, but the intermediate polarizer adds a factor of $\cos^2(60°) = \frac{1}{4}$ to the surviving fraction.

Explanation: When light encounters a series of polarizers, you must apply Malus's Law sequentially at each polarizer — not just at the first and last. Malus's Law states that when polarized light of intensity

I

hits a polarizer at angle

\theta

relative to the light's polarization axis, the transmitted intensity is

I\cos^2\theta

. Here's how to work through this problem step by step. Unpolarized light through the first polarizer always yields

\frac{I_0}{2}

, regardless of orientation. The second polarizer is 30° away from the first, so applying Malus's Law:

\frac{I_0}{2}\cos^2(30°) = \frac{I_0}{2} \cdot \frac{3}{4} = \frac{3I_0}{8}

. The third polarizer is 60° away from the second (since it's at 90° and the second is at 30°), giving:

\frac{3I_0}{8}\cos^2(60°) = \frac{3I_0}{8} \cdot \frac{1}{4} = \frac{3I_0}{32}

. Choice A is correct. Choice B incorrectly applies

\cos^2(90°)

to the full span first, then multiplies by

\cos^2(30°)

— this scrambles the sequential logic and treats the angles non-independently. Choice C reflects a common misconception: yes, two crossed polarizers alone block all light, but an intermediate polarizer at a non-zero, non-90° angle rotates the polarization state, allowing some light through. Choice D correctly identifies the

\frac{I_0}{2}

starting intensity but then misapplies

\cos^2(90°)

across the full span, ignoring the intermediate step entirely. Your key takeaway: always chain Malus's Law step-by-step using the angle between consecutive polarizers, not the total angle from first to last.

Question 3

A quarter-wave plate (QWP) has its fast axis oriented at $45°$ to the horizontal. Horizontally polarized light (electric field along the x-axis) enters the QWP.

What is the polarization state of the light emerging from the QWP, and what happens if this emerging light then passes through a second QWP with its fast axis also at $45°$ to the horizontal?

The light emerging from the first QWP is circularly polarized; after the second QWP (identical orientation), the light becomes linearly polarized but oriented at $90°$ to the original polarization direction (vertical). (correct answer)
The light emerging from the first QWP is elliptically polarized; after the second identical QWP, the light returns to horizontal linear polarization because each QWP undoes the effect of the other.
The light emerging from the first QWP is circularly polarized; after the second identical QWP, the light becomes linearly polarized and returns to its original horizontal orientation because two quarter-wave retardations cancel.
The light emerging from the first QWP is linearly polarized at $45°$ ; after the second identical QWP, the light becomes circularly polarized because a QWP converts $45°$ -polarized light to circular polarization.

Explanation: When working with wave plates, the key is tracking the phase shift each optical element introduces between the fast and slow axis components of the electric field. A quarter-wave plate introduces a

\frac{\lambda}{4}

(or

90°

) phase retardation between components. When horizontally polarized light enters a QWP with its fast axis at

45°

, the field decomposes equally into fast-axis and slow-axis components. After passing through, one component is retarded by

90°

relative to the other, producing two equal-amplitude components with a

90°

phase difference — the definition of circular polarization. This confirms the first part of answer A. Now, when this circularly polarized light enters the second identical QWP (same

45°

orientation), another

90°

phase retardation is added. The total retardation becomes

180°

, which converts circular polarization back into linear polarization. However, the cumulative

180°

phase shift effectively rotates the linear polarization by

90°

from the original — yielding vertically polarized light. Answer A is correct. Answer B is wrong on two counts: the first QWP produces circular (not elliptical) polarization, and the second QWP does not undo the first — it adds another

90°

shift. Answer C is wrong because two identical

90°

retardations sum to

180°

, rotating the polarization to vertical, not returning it to horizontal. Answer D is wrong from the start — horizontally polarized light entering a

45°

-oriented QWP does not exit linearly polarized at

45°

; it exits circularly polarized. Study tip: Always decompose the input field along the wave plate axes and track cumulative phase shifts step by step — never assume two identical elements "cancel."

Question 4

A beam of linearly polarized light passes through a birefringent crystal of thickness $d$ with ordinary refractive index $n_o$ and extraordinary refractive index $n_e$ . The polarization axis of the incident light makes a $45°$ angle with the optic axis of the crystal. The wavelength of the light in vacuum is $\lambda$ .

After exiting the crystal, the relative phase difference between the two components is $\delta = \frac{2\pi d (n_e - n_o)}{\lambda}$ . If $\delta = \frac{\pi}{4}$ and the light then passes through a linear polarizer whose axis is perpendicular to the original polarization direction (i.e., at $-45°$ to the optic axis, or equivalently $135°$ from horizontal), what fraction of the original intensity is transmitted through the polarizer?

$\dfrac{1}{2}$ , because the crystal splits the beam into two equal-amplitude components and a polarizer perpendicular to the original axis always transmits exactly half the total intensity regardless of phase difference.
$\cos^2\dfrac{\pi}{4} = \dfrac{1}{2}$ , because Malus's law applies and the phase difference acts like a rotation of the polarization by $\delta/2 = \pi/8$ , giving a projection onto the perpendicular axis.
$\dfrac{1}{2}\left(1 - \cos\dfrac{\pi}{4}\right) = \dfrac{1}{2}\left(1 - \dfrac{\sqrt{2}}{2}\right) \approx 0.146$ , because the two components projected onto the perpendicular axis interfere destructively, with the phase difference determining the degree of cancellation. (correct answer)
$0$ , because the final polarizer is perpendicular to the original polarization direction, and a birefringent crystal cannot redirect energy into the blocked polarization channel regardless of the phase retardation introduced.

Explanation: When polarized light enters a birefringent crystal with its polarization at 45° to the optic axis, the crystal decomposes it into two equal-amplitude components: one along the ordinary axis, one along the extraordinary axis. These travel at different speeds, accumulating a phase difference

\delta

. After exiting, the light is no longer simply linearly polarized — its polarization state depends entirely on

\delta

. This is the core concept being tested. To find the transmitted intensity through a polarizer at 135° (perpendicular to the original 45° polarization), you project both components onto that axis and allow them to interfere. With equal amplitudes

E_0/\sqrt{2}

and a phase difference

\delta

, the transmitted amplitude is proportional to

\sin(\delta/2)

, and the transmitted intensity fraction is:

I/I_0 = \tfrac{1}{2}(1 - \cos\delta)

With

\delta = \pi/4

\tfrac{1}{2}(1 - \cos\frac{\pi}{4}) = \tfrac{1}{2}(1 - \frac{\sqrt{2}}{2}) \approx 0.146

. That confirms C. A is wrong because the phase difference absolutely matters — a perpendicular polarizer does not automatically transmit half the intensity. That would only be true for a quarter-wave plate (

\delta = \pi/2

). B misapplies Malus's law: phase retardation is not equivalent to a rotation of polarization direction. Malus's law applies to linearly polarized light hitting a polarizer, not to the output of a birefringent crystal. D is wrong because birefringence can couple energy into the perpendicular polarization channel — that's exactly the purpose of wave plates. Study tip: Whenever a problem combines a birefringent crystal with a downstream polarizer, always write out the Jones vector after the crystal and project it explicitly. Never assume the output is still linearly polarized.

Question 5

A Faraday rotator is a magneto-optic device that rotates the plane of polarization of light by an angle $\theta_F = VBd$ , where $V$ is the Verdet constant, $B$ is the applied magnetic field, and $d$ is the path length. Unlike a birefringent wave plate, a Faraday rotator is non-reciprocal. A beam of linearly polarized light passes forward through a Faraday rotator (rotation $+45°$ ), reflects off a mirror, and passes back through the same Faraday rotator. What is the polarization state of the returning beam relative to the original, and what is the fundamental physical reason for this behavior?

The returning beam is polarized at $+90°$ from the original, because the mirror introduces a $+45°$ phase shift that adds to the Faraday rotation, and the second pass contributes nothing since the beam is now traveling antiparallel to the field.
The returning beam is polarized at $0°$ (same as original), because the reflection at the mirror reverses the handedness of the rotation, so the second pass through the rotator exactly cancels the first $+45°$ rotation.
The returning beam is polarized at $+90°$ from the original, because the Faraday rotation does not reverse when the propagation direction reverses — both passes contribute $+45°$ — due to the time-reversal-breaking nature of the magnetic field interaction. (correct answer)
The returning beam is polarized at $0°$ (same as original), because the Faraday rotation is reciprocal like birefringence — the second pass through the rotator undoes the first rotation, restoring the original polarization direction.

Explanation: When you encounter questions about non-reciprocal optical devices, the key is understanding what "non-reciprocal" physically means: the device behaves differently depending on the direction of travel relative to a fixed external field — not relative to the beam itself. In a Faraday rotator, the rotation direction is determined by the magnetic field's orientation in the lab frame, not by which way light is traveling. When light passes forward through a

+45°

rotator, the polarization rotates

+45°

. After reflecting off the mirror, the beam reverses direction — but the magnetic field doesn't. So the second pass also rotates the polarization by

+45°

in the same absolute sense. The total rotation is

+45° + 45° = +90°

, making answer C correct. Answer A is wrong on two counts: mirrors do not introduce a

+45°

phase shift to polarization angle, and the second pass absolutely does contribute rotation — the field still acts on the returning beam. Answer B describes what would happen with a reciprocal device like a wave plate. In reciprocal optics, reversing propagation undoes the effect. But Faraday rotation breaks time-reversal symmetry, so this cancellation does not occur. Answer D makes the same reciprocity error as B, incorrectly treating the Faraday rotator as equivalent to birefringence. The entire point of the device is that it is not reciprocal — this is exactly why it's used in optical isolators. Study tip: Remember that "non-reciprocal" = "time-reversal symmetry broken by an external magnetic field." Any device exploiting this property (Faraday rotators, optical isolators) will accumulate rotation on round trips rather than canceling it.

Question 6

An experimenter uses a single linear polarizer to test whether an unknown beam of light is (i) unpolarized, (ii) partially polarized, or (iii) completely circularly polarized. She rotates the polarizer through a full $360°$ and measures the transmitted intensity as a function of angle. Which of the following correctly distinguishes all three cases based solely on this measurement?

Case (i) and case (iii) both show constant transmitted intensity regardless of polarizer angle, making them indistinguishable by this method alone; case (ii) shows a sinusoidal variation with a nonzero minimum, distinguishing it from the other two. (correct answer)
Case (i) shows constant transmitted intensity, case (iii) shows intensity varying as $\cos^2\theta$ with a zero minimum, and case (ii) shows sinusoidal variation with a nonzero minimum, so all three are distinguishable.
Case (iii) shows constant transmitted intensity, but cases (i) and (ii) both show sinusoidal variation; case (i) has a zero minimum while case (ii) has a nonzero minimum, making case (iii) indistinguishable from case (i).
All three cases produce sinusoidal intensity variation, but the amplitude of the sinusoid differs: zero amplitude for case (i), maximum amplitude for case (iii), and intermediate amplitude for case (ii), so all three are distinguishable.

Explanation: When analyzing polarized light with a single linear polarizer, the key concept is how each light type responds to Malus's Law — and crucially, which cases are experimentally indistinguishable. Unpolarized light has electric field components distributed equally in all directions. A linear polarizer always transmits exactly half the intensity regardless of orientation, giving constant output. Circularly polarized light rotates its electric field vector continuously, but at any instant its amplitude is equal in all directions — so a linear polarizer also transmits exactly half the intensity at every angle, again producing constant output. Both cases are therefore indistinguishable by this measurement alone. Partially polarized light, however, has a preferred direction with unequal intensity distribution. As you rotate the polarizer, transmitted intensity follows a sinusoidal pattern oscillating between a nonzero maximum and a nonzero minimum — it never drops to zero because it retains an unpolarized component. This makes A correct: cases (i) and (iii) both yield constant transmitted intensity, making them indistinguishable, while case (ii) shows sinusoidal variation with a nonzero minimum. B is wrong because it claims circularly polarized light behaves like linearly polarized light (

\cos^2\theta

with a zero minimum) — this confuses circular polarization with linear polarization. C incorrectly swaps which cases are indistinguishable, claiming unpolarized and partially polarized both show sinusoidal behavior. D is entirely wrong — neither unpolarized nor circularly polarized light produces any sinusoidal variation at all. Study tip: Remember that circular polarization and unpolarized light are the classic "impostor pair" — both produce constant intensity through a rotating linear polarizer, and only a wave plate followed by a polarizer can distinguish them.

Question 7

Light scattered from the atmosphere at exactly $90°$ from the direction of the incident sunlight is observed to be completely linearly polarized. A student claims this proves that sunlight itself must be polarized before it enters the atmosphere. Which of the following best evaluates this claim?

The claim is partially correct. Unpolarized incident light produces partially polarized scattered light at $90°$ , but complete polarization at $90°$ requires a small degree of pre-existing polarization in the incident sunlight to suppress the remaining unpolarized component.
The claim is correct. Only if the incident light contains a linearly polarized component can the scattering process selectively transmit one polarization at $90°$ ; unpolarized incident light would produce unpolarized scattered light at all angles.
The claim is incorrect. Complete polarization at $90°$ occurs because the scattering medium acts as a birefringent material, splitting unpolarized light into two orthogonally polarized rays and absorbing one of them regardless of the incident polarization state.
The claim is incorrect. Rayleigh scattering of unpolarized light at $90°$ produces completely linearly polarized scattered light because the oscillating dipoles induced in gas molecules cannot radiate in the direction of their own oscillation, eliminating the component parallel to the scattering plane. (correct answer)

Explanation: Whenever you see a question about atmospheric polarization, anchor your thinking to Rayleigh scattering mechanics — specifically, how induced dipoles radiate. When sunlight strikes a gas molecule, it drives electrons into oscillation, creating an electric dipole. That dipole radiates like a tiny antenna: it emits strongly perpendicular to its oscillation axis and not at all along that axis. Now consider unpolarized sunlight, which you can decompose into two orthogonal polarization components. Each component drives dipoles along its own direction. When you observe scattered light at exactly

90°

from the incident beam, only the component whose dipole oscillates perpendicular to your line of sight can reach you — the component whose dipole points toward you is completely suppressed (a dipole radiates zero intensity along its own axis). This geometric filtering is total and automatic, requiring no pre-existing polarization in the sunlight. The result is complete linear polarization at

90°

, making D correct. A is wrong because it inverts the logic — complete polarization at

90°

doesn't require pre-polarized incident light; it arises naturally from the dipole radiation pattern. B is wrong on a more fundamental level: it incorrectly claims unpolarized light would scatter unpolarized light at all angles, which contradicts well-established Rayleigh scattering theory. C introduces birefringence and absorption, which describe anisotropic crystals (like calcite), not gas molecules in the atmosphere — this is a different physical mechanism entirely and doesn't apply here. Your study tip: remember that Rayleigh scattering at $90°$ is a geometry problem, not a polarization-of-the-source problem. The dipole simply cannot radiate toward you along its own axis.

Question 8

Two coherent beams of left-circularly polarized (LCP) and right-circularly polarized (RCP) light of equal amplitude $E_0$ are superposed. The RCP beam has an additional phase advance of $\phi$ relative to the LCP beam.

What is the polarization state of the superposed beam, and at what angle from the reference axis is the electric field oscillating?

The superposed beam is linearly polarized, oscillating at an angle of $\phi$ from the reference axis, because the phase advance directly sets the polarization angle through the relationship $\theta = \phi$ when the two circular components have equal amplitude.
The superposed beam is linearly polarized, oscillating at an angle of $\phi/2$ from the reference axis, because the superposition of equal-amplitude LCP and RCP with relative phase $\phi$ always produces linear polarization whose orientation depends on the phase difference. (correct answer)
The superposed beam is elliptically polarized with the major axis at $\phi/2$ , because equal-amplitude circular components of opposite handedness produce elliptical polarization whenever a nonzero phase difference exists between them.
The superposed beam is linearly polarized at angle $\phi/2$ only when $\phi$ is a multiple of $\pi$ ; for other values the beam is elliptically polarized, because the phase difference distorts the circular symmetry of each component unequally.

Explanation: When you superpose left- and right-circularly polarized light, think of it as the inverse of circular decomposition: any linearly polarized wave can be built from equal-amplitude LCP and RCP components, so combining them should recover linear polarization. To see why mathematically, write LCP and RCP in component form. LCP goes as

(\hat{x}\cos\omega t + \hat{y}\sin\omega t)

and RCP as

(\hat{x}\cos(\omega t+\phi) - \hat{y}\sin(\omega t+\phi))

. Adding them and applying sum-to-product identities, the

\hat{x}

component becomes

2E_0\cos(\phi/2)\cos(\omega t + \phi/2)

and the

\hat{y}

component becomes

2E_0\cos(\phi/2)\sin(\phi/2 - \omega t + \omega t)\cdots

, which ultimately collapses to a single linear oscillation at angle

\theta = \phi/2

from the reference axis. The result is always linearly polarized regardless of the value of

\phi

, confirming B. A is wrong because it claims

\theta = \phi

, off by a factor of two. This is a classic algebraic error — forgetting that the phase difference splits evenly between the two components when you extract the resultant angle. C is wrong because it incorrectly states equal-amplitude opposite-handedness components produce elliptical polarization. Elliptical polarization arises when the two circular components have unequal amplitudes; equal amplitudes always give linear. D is wrong for the same reason as C — it invents a conditional where none exists. The linearity of the result holds for all values of

\phi

, not just multiples of

\pi

. Study tip: Remember the factor of two — a phase difference

\phi

between circular components rotates the linear polarization by

\phi/2

, not

\phi

Question 9

A half-wave plate (HWP) has its fast axis oriented at angle $\alpha$ from the horizontal. A beam of light that is linearly polarized at angle $\theta$ from the horizontal passes through this HWP.

What is the polarization angle of the transmitted light relative to the horizontal, and what value of $\alpha$ would rotate the polarization from $20°$ to $60°$ ?

The transmitted polarization angle is $\alpha - \theta$ ; to rotate from $20°$ to $60°$ , the fast axis must be set to $\alpha = 80°$ because a HWP reflects the polarization direction about the fast axis.
The transmitted polarization angle is $\theta + 2\alpha$ ; to rotate from $20°$ to $60°$ , the fast axis must be set to $\alpha = 20°$ .
The transmitted polarization angle is $2\alpha - \theta$ ; to rotate from $20°$ to $60°$ , the fast axis must be set to $\alpha = 80°$ because the HWP rotates the polarization by the same angle as the fast axis.
The transmitted polarization angle is $2\alpha - \theta$ ; to rotate from $20°$ to $60°$ , the fast axis must be set to $\alpha = 40°$ . (correct answer)

Explanation: When light passes through a half-wave plate, the HWP reflects the polarization angle about its fast axis — it doesn't simply add or subtract a fixed rotation. Visualizing this geometrically is key: if the fast axis sits at angle

\alpha

and your incoming polarization is at

\theta

, the angular "distance" from

\theta

\alpha

\alpha - \theta

, and the outgoing beam lands the same distance past the fast axis. That gives an output angle of

\alpha + (\alpha - \theta) = 2\alpha - \theta

. For the numerical part, you want the output to equal

60°

with input

\theta = 20°

. Setting

2\alpha - 20° = 60°

gives

2\alpha = 80°

, so

\alpha = 40°

. That's exactly what D states, making it the correct answer. A gets the formula completely wrong —

\alpha - \theta

describes neither a reflection nor a rotation; it's a simple difference with no physical basis here. The claim about

\alpha = 80°

is also inconsistent with even its own formula. B uses

\theta + 2\alpha

, which would mean the fast axis adds to the polarization angle regardless of the input direction — that's not how reflection about an axis works, and plugging in numbers gives

20° + 2(20°) = 60°

, which happens numerically but uses the wrong formula for the wrong reason. C has the correct formula

2\alpha - \theta

but then incorrectly solves for

\alpha

, arriving at

80°

instead of

40°

— a straightforward algebra error. Study tip: Memorize the HWP output rule as "reflect

\theta

about

\alpha

," which always yields

2\alpha - \theta

. Then solving for

\alpha

is just simple algebra — don't let the geometry intimidate you into skipping the arithmetic check.

Question 10

Light traveling in air strikes the flat surface of a glass medium ( $n = 1.52$ ) at an angle such that the reflected beam is completely linearly polarized. A physicist then immerses the entire setup in a liquid with refractive index $n_L = 1.20$ . Which of the following correctly describes the new Brewster angle and the polarization state of the reflected light at that new angle?

The new Brewster angle satisfies $\tan\theta_B' = \frac{1.52}{1.20}$ , and at this angle the reflected light is completely linearly polarized with its electric field perpendicular to the plane of incidence. (correct answer)
The Brewster angle is unchanged at $\arctan(1.52)$ because it depends only on the refractive index of the glass, not on the medium of incidence, and the reflected light remains completely polarized.
The new Brewster angle satisfies $\tan\theta_B' = \frac{1.20}{1.52}$ , and at this angle the reflected light is completely linearly polarized with its electric field parallel to the plane of incidence.
The new Brewster angle satisfies $\tan\theta_B' = \frac{1.52}{1.20}$ , and at this angle the reflected light is partially polarized because immersion in a denser medium reduces the polarizing efficiency of the interface.

Explanation: Whenever you see a question involving Brewster's angle, remember that it arises from the ratio of refractive indices across an interface — specifically the ratio of the incident medium to the transmitted medium. The general formula is

\tan\theta_B = \frac{n_t}{n_i}

, where

n_i

is the index of the incident medium and

n_t

is the index of the refracting medium. In air,

n_i = 1.00

, so

\tan\theta_B = 1.52

. Once you submerge the setup in liquid (

n_L = 1.20

), the incident medium changes, giving

\tan\theta_B' = \frac{1.52}{1.20} \approx 1.267

, which is a smaller angle. At this new Brewster angle, the reflected beam is still completely linearly polarized, with its electric field oriented perpendicular to the plane of incidence — that is the defining physical outcome of Brewster's condition, regardless of what media are involved. This confirms A as correct. Choice B is wrong because Brewster's angle absolutely depends on the incident medium. Changing

n_i

from 1.00 to 1.20 changes the ratio and therefore the angle — you cannot ignore the surrounding medium. Choice C has the ratio inverted. Writing

\frac{1.20}{1.52}

would give a ratio less than 1, implying light going from a denser to a rarer medium, which is backwards for this setup. The transmitted (glass) index always goes in the numerator. Choice D gets the formula right but invents a false physical claim. Immersion in a liquid does not reduce polarization efficiency — at any Brewster angle, the reflected light is fully polarized, period. Your study tip: always write Brewster's law as

\tan\theta_B = n_t / n_i

and identify both media carefully. The ratio flips completely if you swap which side is incident.