Optics

Q1 (10 marks)

Use $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ with real-is-positive magnitudes (image on screen ⇒ real).

(a) Original: object $u$, image $v$, $\frac1v+\frac1u=\frac1{20}$. (1) Object moved further out: $u'=u+10$, image moves closer: $v'=v-15$. (1) So $\frac{1}{v-15}+\frac{1}{u+10}=\frac1{20}$. (1)

From config 1: $\frac1v=\frac1{20}-\frac1u=\frac{u-20}{20u}\Rightarrow v=\frac{20u}{u-20}$. (2)

Substitute into config 2 and solve. Try $u=60$: $v=\frac{1200}{40}=30$. Check config 2: $u'=70,\ v'=15$: $\frac1{15}+\frac1{70}=0.0667+0.0143=0.0810\neq0.05$. Reject.

Solve properly: with $v=\frac{20u}{u-20}$, $\frac{1}{\frac{20u}{u-20}-15}+\frac{1}{u+10}=\frac{1}{20}$. $\frac{20u}{u-20}-15=\frac{20u-15(u-20)}{u-20}=\frac{5u+300}{u-20}$. So term $=\frac{u-20}{5u+300}$. Equation: $\frac{u-20}{5u+300}+\frac1{u+10}=\frac1{20}$. (2)

Multiply out. Let me solve: $\frac{u-20}{5(u+60)}+\frac{1}{u+10}=\frac1{20}$. Multiply by $20$: $\frac{4(u-20)}{u+60}+\frac{20}{u+10}=1$. $4(u-20)(u+10)+20(u+60)=(u+60)(u+10)$. $4(u^2-10u-200)+20u+1200=u^2+70u+600$. $4u^2-40u-800+20u+1200=u^2+70u+600$. $4u^2-20u+400=u^2+70u+600$. $3u^2-90u-200=0$. $u=\frac{90\pm\sqrt{8100+2400}}{6}=\frac{90\pm\sqrt{10500}}{6}=\frac{90\pm102.47}{6}$. $u=32.08\ \text{cm}$ (positive root). (1)

(b) $v=\frac{20\times32.08}{12.08}=53.1\ \text{cm}$. Magnification $m=-\frac{v}{u}=-\frac{53.1}{32.08}=-1.66$; magnitude $1.66$, inverted (real image). (3)

Q2 (12 marks)

(a) Critical angle at glass–liquid interface: $\sin\theta_c=\frac{n_L}{n_g}$ (light going from denser glass to liquid, needs $n_L<n_g$). (2) TIR at $42^\circ$ ⇒ $42^\circ\ge\theta_c\Rightarrow \sin42^\circ\ge\frac{n_L}{n_g}$. (1) No TIR at $38^\circ$ ⇒ $38^\circ<\theta_c\Rightarrow \sin38^\circ<\frac{n_L}{n_g}$. (1) So $n_g\sin38^\circ<n_L\le n_g\sin42^\circ$. (1) $1.5\times0.6157=0.9235$; $1.5\times0.6691=1.0037$. (1) Since $n_L\le n_g$ physically required, range: $\boxed{0.924<n_L<1.00}$ (upper bound also $\le n_g$; effective $0.924<n_L\lesssim1.00$). (1)

(b) Soap film in air: reflection has one $\pi$ phase change (air→film) ⇒ bright condition $2n_Lt=(m+\tfrac12)\lambda$. (2) $t=\frac{(m+\frac12)\lambda}{2n_L}$. $m=0$: $t=\frac{0.5\times600}{2\times1.33}=\frac{300}{2.66}=112.8\ \text{nm}$. (2) $m=1$: $t=\frac{1.5\times600}{2.66}=338.3\ \text{nm}$. (1)

Q3 (10 marks)

(a) Shift of central fringe by inserting sheet: path added $=(n-1)t$. This equals shift to 8th fringe position ⇒ $(n-1)t=8\lambda$. (2) $t=\frac{8\lambda}{n-1}=\frac{8\times500\times10^{-9}}{0.60}=6.67\times10^{-6}\ \text{m}=6.67\ \mu\text{m}$. (3)

(b) In water, wavelength $\lambda'=\lambda/n_w$. Fringe width $\beta=\frac{\lambda' D}{d}=\frac{\lambda D}{n_w d}$. (2) $\beta_{air}=\frac{500\times10^{-9}\times1.5}{0.25\times10^{-3}}=3.0\times10^{-3}\ \text{m}=3.0\ \text{mm}$. $\beta_{water}=3.0/1.33=2.26\ \text{mm}$. (2) Physically: wavelength in water is shorter, so fringes are more closely spaced (smaller $\beta$). (1)

Q4 (10 marks)

(a) Angular magnification $M=f_o/f_e=120/2.0=60$. (1.5) Tube length (normal adjustment) $L=f_o+f_e=122\ \text{cm}$. (1.5)

(b) Rayleigh: $\theta_{min}=1.22\lambda/D=\frac{1.22\times550\times10^{-9}}{0.10}=6.71\times10^{-6}\ \text{rad}$. (3) Distance $r=4\times9.46\times10^{15}=3.784\times10^{16}\ \text{m}$. (1) Linear separation $s=r\theta_{min}=3.784\times10^{16}\times6.71\times10^{-6}=2.54\times10^{11}\ \text{m}$. (3) (≈ $1.7$ AU.)

Q5 (8 marks)

(a) After polaroid 1: $I_1=I_0/2$. (1) After 2 (angle $\theta$ from 1): $I_2=\frac{I_0}{2}\cos^2\theta$. (1) Polaroid 3 at $90^\circ$ from 1 ⇒ angle between 2 and 3 is $(90^\circ-\theta)$: $I_3=I_2\cos^2(90^\circ-\theta)=\frac{I_0}{2}\cos^2\theta\sin^2\theta$. (2)

(b) $I_3=\frac{I_0}{2}\cos^2\theta\sin^2\theta=\frac{I_0}{8}\sin^2 2\theta$. Max when $\sin2\theta=1\Rightarrow \theta=45^\circ$. (2) $I_{max}=I_0/8$. (2)

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{"claim":"Q1 quadratic gives u=32.08cm","code":"u=symbols('u',positive=True); sol=solve(3*u**2-90*u-200,u); val=[s for s in sol if s>0][0]; result=abs(float(val)-32.08)<0.1"},
{"claim":"Q2b m=0 thickness 112.8nm","code":"t=(0.5*600)/(2*1.33); result=abs(t-112.78)<0.5"},
{"claim":"Q3a t=6.67 microns","code":"t=8*500e-9/0.60; result=abs(t-6.667e-6)<1e-8"},
{"claim":"Q4b linear separation approx 2.54e11 m","code":"th=1.22*550e-9/0.10; r=4*9.46e15; s=r*th; result=abs(s-2.54e11)/2.54e11<0.02"},
{"claim":"Q5 max intensity I0/8","code":"th=pi/4; frac=Rational(1,2)*cos(th)**2*sin(th)**2; result=simplify(frac-Rational(1,8))==0"}
]