Optics

Question 1 (10 marks)

Setup (3): Source $A=(0,h_1)$ in medium $n_1$, observer $B=(L,-h_2)$ in $n_2$, interface at $y=0$, ray crosses at $(x,0)$. Optical path (∝ time): $L(x) = n_1\sqrt{x^2+h_1^2} + n_2\sqrt{(L-x)^2+h_2^2}.$

Minimize (4): $\frac{dL}{dx} = n_1\frac{x}{\sqrt{x^2+h_1^2}} - n_2\frac{(L-x)}{\sqrt{(L-x)^2+h_2^2}} = 0.$

Interpret (3): The ratios are sines of the angles from the normal: $\sin\theta_1 = \frac{x}{\sqrt{x^2+h_1^2}}, \quad \sin\theta_2 = \frac{L-x}{\sqrt{(L-x)^2+h_2^2}}.$ $\Rightarrow n_1\sin\theta_1 = n_2\sin\theta_2. \qquad\blacksquare$

Question 2 (12 marks)

(a) [4] At critical angle refraction angle $=90°$; Snell: $n_1\sin\theta_c = n_2\sin90° = n_2$, so $\boxed{\theta_c = \sin^{-1}(n_2/n_1)}.$

(b) [3] $\sin\theta_c = 1.46/1.50 = 0.97\overline{3}$, $\theta_c = \sin^{-1}(0.9733) = 76.7°$.

(c) [5] Ray enters air→core at angle $\theta_i$, refracts to $\theta_r$; to guide, the internal angle at the wall $=90°-\theta_r \ge \theta_c$. $n_0\sin\theta_i = n_1\sin\theta_r = n_1\cos(90°-\theta_r)=n_1\cos\theta_{\text{wall,max}} = n_1\cos\theta_c$. $\cos\theta_c = \sqrt{1-(n_2/n_1)^2}$, so with $n_0=1$: $\text{NA}=\sin\theta_{i,\max} = n_1\sqrt{1-(n_2/n_1)^2} = \sqrt{n_1^2-n_2^2}.$ Value: $\sqrt{1.50^2-1.46^2}=\sqrt{2.25-2.1316}=\sqrt{0.1184}=0.344.$

Question 3 (12 marks)

(a) [8] Path difference $\Delta = d\sin\theta$ (2). For bright fringe $\Delta = m\lambda$ (1). Small angle: $\sin\theta\approx\tan\theta = y/D$ (2), so $y_m = m\lambda D/d$ (1). Fringe width $\beta = y_{m+1}-y_m = \lambda D/d$ (2).

(b) [2] $\beta = \frac{589\times10^{-9}\times1.5}{0.20\times10^{-3}} = 4.42\times10^{-3}\,\text{m} = 4.42\,\text{mm}$.

(c) [2] In water $\lambda' = \lambda/n$, so $\beta' = \beta/1.33 = 4.42/1.33 = 3.32\,\text{mm}$.

Question 4 (10 marks)

(a) [5] $\frac1f=(n-1)\left(\frac1{R_1}-\frac1{R_2}\right)$. Equiconvex: $R_1=+20$, $R_2=-20$ cm. $\frac1f=(0.5)\left(\frac1{20}-\frac1{-20}\right)=0.5\times\frac{2}{20}=0.05\,\text{cm}^{-1}$. $f = 20\,\text{cm} = 0.20\,\text{m}$. Power $P_1=+5.0\,\text{D}$.

(b) [5] In contact: $P = P_1+P_2 = 5.0 + (-2.0) = +3.0\,\text{D}$. $f = 1/P = 0.333\,\text{m} = 33.3\,\text{cm}$. Positive → converging.

Question 5 (10 marks)

(a) [6] Divide slit into strips; strip at position $x$ has phase $\frac{2\pi}{\lambda}x\sin\theta$. Sum amplitude: $E\propto\int_{-a/2}^{a/2} e^{i k x\sin\theta}dx = a\,\frac{\sin\beta}{\beta}, \quad \beta=\frac{\pi a\sin\theta}{\lambda}.$ Intensity $\propto|E|^2 \Rightarrow I=I_0(\sin\beta/\beta)^2$ (4). First minimum: $\beta=\pi\Rightarrow a\sin\theta=\lambda$ (2).

(b) [4] Snippet:

import numpy as np
lam = 1.0; a = 5*lam
th = np.linspace(-np.pi/6, np.pi/6, 1000)
beta = np.pi*a*np.sin(th)/lam
I = np.sinc(beta/np.pi)**2   # np.sinc(x)=sin(pi x)/(pi x)
# plt.plot(np.degrees(th), I)

First minima where $a\sin\theta=\lambda\Rightarrow\sin\theta=1/5$, $\theta=\pm11.54°$. (2 for code, 2 for correct minima)

Question 6 (6 marks)

(a) [4] Reflected ⟂ refracted: $\theta_B + \theta_r = 90°$, so $\theta_r = 90°-\theta_B$. Snell: $n_1\sin\theta_B = n_2\sin\theta_r = n_2\cos\theta_B$. $\Rightarrow \tan\theta_B = n_2/n_1$.

(b) [2] At Brewster's angle the refracted ray direction coincides with the would-be reflected direction's dipole oscillation axis; dipoles in the medium oscillate along the refracted E-field, and cannot radiate along their own axis, so the component polarized in the plane of incidence is suppressed in reflection — the reflected beam is fully polarized perpendicular to the plane of incidence.

[
{"claim":"Critical angle for n1=1.5,n2=1.46 is 76.7 deg","code":"import math; tc=math.degrees(math.asin(1.46/1.5)); result = abs(tc-76.7)<0.2"},
{"claim":"Numerical aperture sqrt(1.5^2-1.46^2)=0.344","code":"import math; na=math.sqrt(1.5**2-1.46**2); result = abs(na-0.344)<0.005"},
{"claim":"Fringe width beta=4.42 mm","code":"b=589e-9*1.5/0.20e-3; result = abs(b*1000-4.42)<0.02"},
{"claim":"Combination power +3D gives f=33.3cm","code":"P=5.0-2.0; f=1/P; result = abs(f*100-33.33)<0.1"},
{"claim":"Single-slit first minimum for a=5lam at 11.54 deg","code":"import math; th=math.degrees(math.asin(1/5)); result = abs(th-11.54)<0.05"}
]