17-Phys-A7 Optics · May 2015

Question 5 of 6: Polarizers and a mica retardation plate

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, May 2015, 98-Phys-A7 Optics. Three hours, closed book, one approved Casio or Sharp calculator. Question 1 is mandatory; the rubric asks for Question 1 plus any four of Questions 2–6 (only the first four questions appearing in the answer book are marked) — all six are worked here, because the set is a study resource.

Reference texts. E. Hecht, Optics, 5th ed.; F. L. Pedrotti, L. M. Pedrotti and L. S. Pedrotti, Introduction to Optics, 3rd ed.

Question 5: Polarizers and a mica retardation plate (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a).

Given. Unpolarized light through 3 ideal polarizers; axis 2 at angle $\theta_1$ from axis 1; axis 3 at angle $\theta_2$ from axis 1 (so axis 3 is at $\theta_2-\theta_1$ from axis 2).

Find. Percentage of the incident (unpolarized) power transmitted.

Approach. The first polarizer always halves unpolarized light (no preferred axis to project onto); every polarizer after that follows Malus's law, $I\to I\cos^2(\Delta)$, applied twice more.

1. After polarizer 1. $I_1=\tfrac12 I_0$ (any polarizer transmits exactly half of unpolarized light).
2. After polarizer 2 (Malus, angle $\theta_1$ from axis 1). $I_2=I_1\cos^2\theta_1=\tfrac12 I_0\cos^2\theta_1$.
3. After polarizer 3 (Malus, angle $\theta_2-\theta_1$ from axis 2). $$I_3=I_2\cos^2(\theta_2-\theta_1)=\tfrac12 I_0\cos^2\theta_1\cos^2(\theta_2-\theta_1).$$

Transmitted percentage:

$$\boxed{\dfrac{I_3}{I_0}\times100\% = 50\%\times\cos^2\theta_1\cos^2(\theta_2-\theta_1)}.$$

Quantity	Result
After polarizer 1	$50\%$ of $I_0$
Transmitted fraction	$50\%\cos^2\theta_1\cos^2(\theta_2-\theta_1)$

Part (b) — unpolarized + circular mixture. A single measurement — rotating an ideal linear polarizer and recording transmitted intensity — cannot distinguish this mixture from pure unpolarized light: circularly polarized light transmits exactly half its power through a linear polarizer at any orientation, identical to unpolarized light, so $I(\phi)=\tfrac12(I_u+I_c)$ is constant in $\phi$ either way.

The fix is to insert a quarter-wave plate before the rotatable polarizer, with its fast axis at $45^\circ$ to some reference. A QWP converts the circular component into fully linear polarized light (at $45^\circ$ to its axes) while leaving unpolarized light unpolarized (a wave plate cannot polarize unpolarized light — it only imposes a fixed relative phase, which is irrelevant when there is no phase correlation to begin with). Now scan the analyzer angle $\phi$ after the QWP:

$$I(\phi)=\tfrac12 I_u + I_c\cos^2(\phi-\phi_0).$$

Record $I_{\max}$ (at $\phi=\phi_0$) and $I_{\min}$ (at $\phi=\phi_0+90^\circ$):

$$I_{\min}=\tfrac12 I_u \ \Rightarrow\ I_u=2I_{\min}, \qquad I_c=I_{\max}-I_{\min}.$$

The total power check $I_0=I_u+I_c=I_{\min}+I_{\max}$ confirms the decomposition is complete and consistent.

Part (c) — mica retardation plate.

Given. $n_1=1.600$, $n_2=1.5950$ (so $\Delta n=0.0050$) along the mica's two in-plane optical axes; crossed polarizers, mica fast axis at $45^\circ$ to both (the standard alignment for maximum transmission); observed colour purple (red $\approx650$ nm and blue $\approx450$ nm transmitted, green $\approx550$ nm is not).

Find. (i) possible mica thicknesses; (ii) colour change on rotating the mica; (iii) colour change on rotating the analyzer.

Approach (i). Between crossed polarizers with the plate at $45^\circ$, the transmitted intensity for a given wavelength is $T(\lambda)=\sin^2(\pi\Delta n\, d/\lambda)$ (a standard Jones-calculus result, re-derived in part (iii) below as the $\psi=90^\circ$ case of a more general formula). "Green is absent" means $T(\lambda_{\text{green}})=0$, i.e. the optical path difference is an exact integer number of green wavelengths.

1. Thickness condition. $T=0\Rightarrow \Delta n\,d=m\lambda_{\text{green}}$ for a positive integer $m$, taking $\lambda_{\text{green}}=550$ nm: $$d_m=\frac{m\lambda_{\text{green}}}{\Delta n}=\frac{m\times550\ \text{nm}}{0.0050}=m\times110\ \mu\text{m}, \quad m=1,2,3,\dots$$ So $\boxed{d=110\ \mu\text{m},\ 220\ \mu\text{m},\ 330\ \mu\text{m},\dots}$ are all "possible thicknesses" — any integer multiple of $110\ \mu$m puts a green minimum at crossed polarizers.
2. Confirm the colour is plausibly purple at the thinnest sheet ($m=1$, $d=110\ \mu$m, $\Delta n\,d=550$ nm). $$\begin{gathered} T(\lambda_r)=\sin^2\!\Big(\frac{\pi\times550}{650}\Big)=0.216 \\ T(\lambda_b)=\sin^2\!\Big(\frac{\pi\times550}{450}\Big)=0.413 \\ T(\lambda_g)=0 \end{gathered}$$ Substantial red and blue with zero green is exactly the description of purple/magenta given in the question — and $d\approx110\ \mu$m ($\Delta n\,d\approx550$ nm retardation) is precisely the mineralogists' "sensitive-tint" (full-wave, first-order red/magenta) plate, a well-known real mica compensator thickness, which is a good independent check on the arithmetic.

Approach (ii) — rotating the mica (polarizers fixed, crossed). Rotating the plate's fast axis to angle $\phi$ from the polarizer scales every wavelength's transmission by the same $\sin^2(2\phi)$ factor (standard retarder-between-crossed-polarizers result), on top of the spectral shape $\sin^2(\pi\Delta n\,d/\lambda)$ found above:

$$T(\lambda,\phi)=\sin^2(2\phi)\,\sin^2\!\Big(\frac{\pi\Delta n\,d}{\lambda}\Big).$$

Since the $\sin^2(2\phi)$ factor is the same number for every $\lambda$, it scales overall brightness only — the hue (purple) does not change as the mica rotates. The intensity, however, goes through four extinctions per revolution (complete darkness at $\phi=0^\circ,90^\circ,180^\circ,270^\circ$, where the mica axes align with a polarizer and it acts as if absent) and four brightness maxima at $\phi=45^\circ,135^\circ,225^\circ,315^\circ$ (the orientation used above).

Approach (iii) — rotating the analyzer (polarizer and mica fixed at $45^\circ$). Using Jones calculus with the polarizer along $x$, the mica (fast axis $45^\circ$, retardance $\Gamma=2\pi\Delta n\,d/\lambda$) next, and an analyzer at angle $\psi$ from the polarizer:

$$T(\psi)=\tfrac12\big[1+\cos(2\psi)\cos\Gamma\big].$$

1. Recover the crossed case. At $\psi=90^\circ$: $T=\tfrac12(1-\cos\Gamma)=\sin^2(\Gamma/2)=\sin^2(\pi\Delta n\,d/\lambda)$ — matches part (i) exactly.
2. Parallel analyzer, $\psi=0^\circ$. $T=\tfrac12(1+\cos\Gamma)=\cos^2(\pi\Delta n\,d/\lambda)$: the exact complement of the crossed spectrum. Green, which was fully blocked at $\psi=90^\circ$ ($\Gamma=2\pi m$, $\cos\Gamma=1$), is now fully transmitted ($T=1$), while red and blue dim — the transmitted colour flips to green, the complementary colour to purple.
3. Analyzer at $45^\circ$ (parallel to the mica axis). $\cos(2\times45^\circ)=0$, so $T=\tfrac12$ for every wavelength — the light is transmitted colourlessly (as white/grey light at half intensity), the midpoint of the colour swing.

So as the analyzer rotates from crossed ($90^\circ$) through $45^\circ$ to parallel ($0^\circ$), the transmitted colour shifts continuously purple $\to$ colourless (white, half intensity) $\to$ green, and (since $T$ depends on $\psi$ only through $\cos2\psi$) the whole cycle repeats every $180^\circ$ of analyzer rotation, with purple recurring at $\psi=90^\circ,270^\circ$ and green at $\psi=0^\circ,180^\circ$.

Quantity	Result
Possible mica thicknesses	$d_m=m\times110\ \mu\text{m}$, $m=1,2,3,\dots$
Rotate mica (crossed polarizers)	hue fixed (purple); intensity $\propto\sin^2(2\phi)$, 4 extinctions/rev
Rotate analyzer ($\psi$ from polarizer)	$T(\psi)=\tfrac12[1+\cos2\psi\cos\Gamma]$: purple (90°) → white (45°) → green (0°)

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