Optics

Question 1

(a) Path lengths: $AP=\sqrt{x^2+h_1^2}$, $PB=\sqrt{(d-x)^2+h_2^2}$. Optical path $L(x)=n_1\sqrt{x^2+h_1^2}+n_2\sqrt{(d-x)^2+h_2^2}$. (2) $\frac{dL}{dx}=\frac{n_1 x}{\sqrt{x^2+h_1^2}}-\frac{n_2(d-x)}{\sqrt{(d-x)^2+h_2^2}}=0.$ (2) Recognise $\dfrac{x}{\sqrt{x^2+h_1^2}}=\sin\theta_1$ and $\dfrac{d-x}{\sqrt{(d-x)^2+h_2^2}}=\sin\theta_2$ (angles from normal). (2) Hence $n_1\sin\theta_1=n_2\sin\theta_2$. (1)

(b) $\frac{d^2L}{dx^2}=\dfrac{n_1 h_1^2}{(x^2+h_1^2)^{3/2}}+\dfrac{n_2 h_2^2}{((d-x)^2+h_2^2)^{3/2}}$. Both terms positive $\Rightarrow L''>0 \Rightarrow$ minimum. (4) (2 for correct derivative, 2 for sign argument.)

(c) At critical angle refracted ray grazes ($\theta_2=90^\circ$): $n_1\sin\theta_c=n_2\sin90^\circ=n_2$, so $\sin\theta_c=\frac{n_2}{n_1},\qquad \theta_c=\arcsin\!\frac{n_2}{n_1}.$ (3) Requires $n_1>n_2$ (dense→rare). (1)

(d) At the core–cladding boundary TIR needs incidence angle $\ge\theta_c$ with $\sin\theta_c=n_{\text{clad}}/n_{\text{core}}$. The refraction angle inside the core (from the end face) $\theta_r$ satisfies $\theta_r=90^\circ-\theta_c$ for the marginal ray. At entry face: $\sin\theta_a=n_{\text{core}}\sin\theta_r=n_{\text{core}}\cos\theta_c$. (2) $\sin\theta_a=n_{\text{core}}\sqrt{1-\frac{n_{\text{clad}}^2}{n_{\text{core}}^2}}=\sqrt{n_{\text{core}}^2-n_{\text{clad}}^2}.$ (2) $=\sqrt{1.50^2-1.45^2}=\sqrt{2.25-2.1025}=\sqrt{0.1475}=0.3841$, so $\theta_a=\arcsin(0.3841)=22.6^\circ$. (1)

Question 2

(a) Apply surface relation at surface 1 (air $n_1=1\to n$): $\dfrac{n}{v_1}-\dfrac{1}{u}=\dfrac{n-1}{R_1}$. At surface 2 ($n\to1$), image of surface 1 acts as object: $\dfrac{1}{v}-\dfrac{n}{v_1}=\dfrac{1-n}{R_2}$. (3) Add: $\frac{1}{v}-\frac{1}{u}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$ With object at $\infty$, $v=f$: $\frac{1}{f}=(n-1)\left(\frac1{R_1}-\frac1{R_2}\right)$. (2) Sign convention: distances measured from lens, +ve in direction of light propagation; $R>0$ if centre of curvature on outgoing side. (1)

(b) $P=P_1+P_2-x P_1 P_2 = 7 - 10x$ (with $x$ in metres). (3) Set $=6$: $10x=1\Rightarrow x=0.10\,\text{m}$. (2)

(c) $\frac1{R_1}-\frac1{R_2}=\frac{1}{0.20}-\frac{1}{-0.20}=5+5=10\,\text{m}^{-1}$. (1) $f_{\text{red}}=\dfrac{1}{(1.514-1)(10)}=\dfrac{1}{5.14}=0.19455\,\text{m}$. (1.5) $f_{\text{violet}}=\dfrac{1}{(1.524-1)(10)}=\dfrac{1}{5.24}=0.19084\,\text{m}$. (1.5) Axial chromatic aberration $f_{\text{red}}-f_{\text{violet}}=0.00371\,\text{m}\approx3.7\,\text{mm}$. (1) Violet has larger $n$ → shorter focal length → focuses nearer the lens. (1)

(d) Combining a crown (low dispersion, positive) and flint (high dispersion, negative) lens: total power $P=P_c+P_f$; the condition $\frac{P_c}{V_c}+\frac{P_f}{V_f}=0$ (V = Abbe number) makes $dP/d\lambda=0$ to first order, so red and violet share a common focus while net power stays positive. (3)

Question 3

(a) Path difference between slits to point $P$: $\Delta=d\sin\theta\approx d\,y/D$ for small $\theta$ ($D\gg d$). (2) Bright fringe: $\Delta=m\lambda\Rightarrow y_m=m\lambda D/d$. (2) Fringe width $\beta=y_{m+1}-y_m=\lambda D/d$. (1)

(b) Divide slit into elements $dx$; contribution $\propto dx\,e^{i k x\sin\theta}$. Amplitude $E\propto\int_{-a/2}^{a/2}e^{ikx\sin\theta}dx=a\,\frac{\sin\beta}{\beta}$, $\beta=\frac{ka\sin\theta}{2}=\frac{\pi a\sin\theta}{\lambda}$. (4) Intensity $I=I_0(\sin\beta/\beta)^2$. (1) Minima where $\sin\beta=0,\ \beta\ne0$: $\beta=m\pi\Rightarrow a\sin\theta=m\lambda$, $m=\pm1,\pm2,...$ (1)

(c) $\beta_{\text{fringe}}=\lambda D/d=\dfrac{600\times10^{-9}\times2.0}{0.50\times10^{-3}}=2.4\times10^{-3}\,\text{m}=2.4\,\text{mm}$. (2) Central diffraction maximum half-angle: $a\sin\theta=\lambda\Rightarrow\sin\theta_1=\lambda/a=6\times10^{-3}$. Interference maxima at $d\sin\theta=m\lambda\Rightarrow \sin\theta=m\lambda/d$. Ratio $d/a=5$: missing orders at $m=\pm5,\pm10,...$ So within $|m|<5$: orders $m=-4..4$ = 9 bright fringes fully within central envelope. (3)

(d) Pseudocode:

import numpy as np
theta = np.linspace(-A, A, N)
s = np.sin(theta)
beta  = pi*a*s/lam
diff  = (np.sinc(beta/pi))**2          # sin(beta)/beta squared
inter = np.cos(pi*d*s/lam)**2
I = diff*inter
# peak find: local maxima
peaks = [i for i in 1..N-2 if I[i]>I[i-1] and I[i]>I[i+1]]

Rayleigh criterion: two sources resolved if the central max of one falls on the first min of the other, i.e. angular separation $\ge\theta_R=1.22\lambda/D$ (aperture $D$). In code: compute combined intensity of two shifted patterns; check for a dip (local minimum) between the two central peaks — if present, resolved. (4)

[
{"claim":"Fibre acceptance angle sqrt(nc^2-ncl^2)=0.3841 -> 22.6deg","code":"import math\nval=math.sqrt(1.50**2-1.45**2)\nang=math.degrees(math.asin(val))\nresult = abs(val-0.3841)<1e-3 and abs(ang-22.6)<0.2"},
{"claim":"Two-lens spacing for P=6: x=0.10 m","code":"from sympy import symbols, solve\nx=symbols('x')\nsol=solve(7-10*x-6,x)\nresult = sol[0]==sympy.Rational(1,10)"},
{"claim":"Chromatic aberration f_red-f_violet approx 3.7 mm","code":"fr=1/((1.514-1)*10)\nfv=1/((1.524-1)*10)\nresult = abs((fr-fv)-0.00371)<5e-5"},
{"claim":"Fringe width 2.4 mm and 9 fringes in central max","code":"beta=600e-9*2.0/0.5e-3\nratio=0.5e-3/0.1e-3\nfringes=2*(int(ratio)-1)+1\nresult = abs(beta-2.4e-3)<1e-6 and fringes==9"}
]