JEE Advanced 2024 Paper 1 · Q16 · Geometrical Optics
A light ray is incident on the surface of a sphere of refractive index $n$ at an angle of incidence $\theta_0$. The ray partially refracts into the sphere with angle of refraction $\phi_0$ and then partly reflects from the back surface. The reflected ray then emerges out of the sphere after a partial refraction. The total angle of deviation of the emergent ray with respect to the incident ray is $\alpha$. Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
List-I: (P) If $n = 2$ and $\alpha = 180^\circ$, then all the possible values of $\theta_0$ will be (Q) If $n = \sqrt{3}$ and $\alpha = 180^\circ$, then all the possible values of $\theta_0$ will be (R) If $n = \sqrt{3}$ and $\alpha = 180^\circ$, then all the possible values of $\phi_0$ will be (S) If $n = \sqrt{2}$ and $\theta_0 = 45^\circ$, then all the possible values of $\alpha$ will be
List-II: (1) $30^\circ$ and $0^\circ$; (2) $60^\circ$ and $0^\circ$; (3) $45^\circ$ and $0^\circ$; (4) $150^\circ$; (5) $0^\circ$.
[Figure: ray incident on sphere, refracted inside, reflects at back surface, refracts out.]
Reveal answer + step-by-step solution
Correct answer:A
Solution
Snell at entry/exit: $\sin\theta_0 = n\sin\phi_0$. The geometry gives total deviation $\alpha = 2(\theta_0 - \phi_0) + (180^\circ - 2\phi_0) = 180^\circ + 2\theta_0 - 4\phi_0$.
(P) $n=2$, $\alpha = 180^\circ$: $2\theta_0 = 4\phi_0 \Rightarrow \theta_0 = 2\phi_0$. With $\sin 2\phi_0 = 2\sin\phi_0$ ⇒ $2\sin\phi_0\cos\phi_0 = 2\sin\phi_0$ ⇒ $\cos\phi_0 = 1$ ⇒ $\phi_0 = 0$, so $\theta_0 = 0^\circ$. → (5) only $0^\circ$.
(Q) $n=\sqrt 3$, $\alpha=180^\circ$: $\theta_0=2\phi_0$ and $\sin 2\phi_0 = \sqrt 3 \sin\phi_0$ ⇒ $\cos\phi_0 = \sqrt 3/2$ ⇒ $\phi_0 = 30^\circ$, $\theta_0 = 60^\circ$. Plus the trivial $\phi_0=\theta_0=0$. → $\theta_0 \in \{0^\circ, 60^\circ\}$ = (2).
(R) Same condition, possible $\phi_0 \in \{0^\circ, 30^\circ\}$ = (1).
(S) $n=\sqrt 2$, $\theta_0=45^\circ$: $\sin\phi_0 = \sin 45^\circ/\sqrt 2 = 1/2$, so $\phi_0 = 30^\circ$. Then $\alpha = 180^\circ + 2(45^\circ) - 4(30^\circ) = 180^\circ + 90^\circ - 120^\circ = 150^\circ$. → (4). Mapping P→5, Q→2, R→1, S→4 = option (A).
Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →