17-Phys-A7 Optics · May 2016

Question 6 of 6: Three polarizers, a quarter-wave plate, and a birefringent retarder

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

Notes on this paper

98-Phys-A7, Optics — National Exams, May 2016. 3 hours; closed book (formula sheet supplied). Notes 1–7 on the cover page require Q1 and Q2 (mandatory) plus any two of Q3–Q6 for a complete paper; every question is solved below as a full study resource, including all four of Q3–Q6.

Reference texts. Hecht, Optics, 5th ed.; Pedrotti, Pedrotti & Pedrotti, Introduction to Optics, 3rd ed.; Griffiths, Introduction to Electrodynamics, 4th ed. (Ch. 9, EM waves in matter).

The 6 real questions (1–6, each 15 marks) are solved in full below.

Question 6: Three polarizers, a quarter-wave plate, and a birefringent retarder (15 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) — three polarizers.

Given. Unpolarized incident light $I_0$; polarizer 2's axis at $\theta_1$ from polarizer 1; polarizer 3's axis at $\theta_2$ from polarizer 1 (so at $\theta_2-\theta_1$ from polarizer 2).

Find. Percentage of $I_0$ transmitted through all three.

Approach. The first (ideal) polarizer halves unpolarized light regardless of its own orientation; each subsequent ideal polarizer applies Malus's law using the angle to the PREVIOUS polarizer's axis (the light is now linearly polarized along that axis).

1. After polarizer 1. $I_1=\tfrac12 I_0$ (unpolarized $\to$ linear, axis 1).
2. After polarizer 2 (Malus, angle $\theta_1$). $I_2=I_1\cos^2\theta_1$.
3. After polarizer 3 (Malus, angle $\theta_2-\theta_1$ from polarizer 2's own axis). $I_3=I_2\cos^2(\theta_2-\theta_1)$, so $$\boxed{\frac{I_3}{I_0}=\frac12\cos^2\theta_1\,\cos^2(\theta_2-\theta_1)}.$$
4. Illustrative numeric check. $\theta_1=25^\circ,\ \theta_2=65^\circ$ (so $\theta_2-\theta_1=40^\circ$): $\cos^225^\circ=0.8214$, $\cos^240^\circ=0.5868$, giving $I_3/I_0=\tfrac12(0.8214)(0.5868)=\boxed{24.1\%}$.

Quantity	Value
General transmittance	$\tfrac12\cos^2\theta_1\cos^2(\theta_2-\theta_1)$
Illustrative case ($\theta_1=25^\circ,\theta_2=65^\circ$)	$24.1\%$

Part (b) — QWP + rotating polarizer extinction. Any FULLY polarized beam can be decomposed, along its own polarization-ellipse principal axes, into two orthogonal linear components with a fixed relative phase of exactly $\pm90^\circ$ (that phase relationship is the definition of an ellipse's principal axes; for circular light the two amplitudes are also equal). A quarter-wave plate adds a further $90^\circ$ of relative phase between whatever two orthogonal components lie along ITS fast/slow axes. If the QWP's axes are aligned with the incident beam's own principal axes, the $\pm90^\circ$ intrinsic phase difference is shifted to $0^\circ$ or $180^\circ$ — i.e. the light emerging from the QWP is linearly polarized, which a rotating polarizer CAN extinguish at one orientation. Since the observer already found that no polarizer angle alone gives zero (ruling out simple linear or unpolarized/partially-polarized light), the incident light must be elliptically (including the special case of circularly) polarized: placing the QWP with its axes along the ellipse's own axes converts it to linear, and the analyzer can then be rotated to block it.

Part (c)(i) — sketch.

Part (c)(ii) — required birefringence.

Given. Linear input at $45^\circ$ to the plate's fast/slow axes (equal-amplitude decomposition, the standard alignment for maximum-purity circular output), minimum thickness $t$, wavelength $\lambda$.

Find. $\Delta n=|n_s-n_f|$ for a minimum-thickness quarter-wave plate.

Approach. Linear $\to$ circular conversion needs the two orthogonal components to emerge with a relative phase of exactly $\pi/2$ (a quarter cycle); the optical path difference for the minimum thickness is one quarter-wave.

1. Quarter-wave condition. $\Delta n\cdot t=\dfrac{\lambda}{4}\ \Rightarrow\ \boxed{\Delta n=\dfrac{\lambda}{4t}}$ (the smallest $\Delta n\cdot t$ that gives a net $90^\circ$ relative phase, since larger integer-plus-quarter multiples also work but $t$ would not be minimum).
2. Illustrative numeric check. For $\lambda=589$ nm (sodium D line) and a representative $t=100\ \mu$m: $\Delta n=\dfrac{589\times10^{-9}}{4(100\times10^{-6})}=\boxed{1.47\times10^{-3}}$ — a realistic mica/quartz-scale birefringence.

Quantity	Value
Minimum-thickness QWP condition	$\Delta n=\lambda/(4t)$
Illustrative ($\lambda=589$ nm, $t=100\ \mu$m)	$\Delta n=1.47\times10^{-3}$

Part (c)(iii) — entering and exiting fields. With the plate's fast axis along $\hat f$ and slow axis along $\hat s$, and the linear input at $45^\circ$ to both (amplitude $E_0$ resolved equally along each):

$$\mathbf E_{\text{in}}(0,t)=\frac{E_0}{\sqrt2}(\hat f+\hat s)\cos(\omega t),$$

and after the slow component accumulates an extra quarter-cycle of phase lag relative to the fast component (the birefringent path difference derived above):

$$\mathbf E_{\text{out}}(L,t)=\frac{E_0}{\sqrt2}\Big[\hat f\cos(\omega t-kL_f)+\hat s\cos\big(\omega t-kL_f-\tfrac\pi2\big)\Big]=\frac{E_0}{\sqrt2}\big[\hat f\cos(\omega t')+\hat s\sin(\omega t')\big],$$

with $t'=t-L_f/v_f$ (an overall, physically irrelevant time/phase offset absorbed for clarity). This is exactly the equal-amplitude, $90^\circ$-quadrature form of a circularly polarized wave: at fixed position, $\mathbf E$ traces a circle of radius $E_0/\sqrt2$ in the $f$-$s$ plane as $t'$ advances, confirming the conversion.

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