A light ray is incident on the surface of a sphere of refractive index $n$ at an angle of incidence…

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.

1. $\mathrm{P} \rightarrow 5 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 4$
2. $\mathrm{P} \rightarrow 5 ; \mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 4$
3. $\mathrm{P} \rightarrow 3 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 4$
4. $\mathrm{P} \rightarrow 3 ; \mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 5$

Solution

$\begin{aligned} & \alpha=\left(\theta_0-\phi_0\right)+\left(180-2 \phi_0\right)+\left(\theta_0-2 \phi_0\right) \\ & \alpha=180+2 \theta_0-4 \phi_0\end{aligned}$ $\textbf{(P) } \alpha=180+2 \theta_0-4 \phi_0$ $180=180+2 \theta_0-4 \phi_0 \Rightarrow \theta_0=2 \phi_0$ ...(i) $\sin \theta_0=2 \sin \phi_0$ ...(ii) From (i) & (ii) $\begin{aligned} & \sin \theta_0=2 \sin \left(\theta_0 / 2\right) \Rightarrow \cos \left(\frac{\theta_0}{2}\right)=1 \\ & \frac{\theta_0}{2}=0 \\ & \Rightarrow \theta_0=0\end{aligned}$ $\textbf{(Q) } \theta_0=2 \phi_0$ ...(i) $\sin \theta_0=\sqrt{3} \sin \phi_0$ ...(ii) From (i) & (ii) $\begin{aligned} & \sin \theta_0=\sqrt{3} \sin \left(\frac{\theta_0}{2}\right) \\ & \Rightarrow \cos \left(\frac{\theta_0}{2}\right)=\frac{\sqrt{3}}{2} \\ & \frac{\theta_0}{2}=30,150 \\ & \theta_0=60,300 \text { (Rejected) } \\ & \theta_0=60 , 0\end{aligned}$ $\textbf{(R) } \theta_0=2 \phi_0$ $\begin{aligned} & \sin \theta_0=\sqrt{3} \sin \phi_0 \\ & \sin 2 \theta_0=\sqrt{3} \sin \phi_0 \\ & \cos \phi_0=\frac{\sqrt{3}}{2} \\ & \phi_0=30,150 \text { (Rejected) }\end{aligned}$ $\phi_0=30, 0$ ...(iii) $\begin{aligned} & \textbf {(S) } \sin 45=\sqrt{2} \cos \phi_0 \\ & \cos \phi_0=1 / 2 \\ & \phi_0=60 \\ & \alpha=180+2 \theta_0-4 \phi_0\end{aligned}$ $\alpha=180+90-120$ ...(iv) $=180-30 ; \alpha=150^{\circ}$ ;

Asked in: JEE Advanced 2024 (Paper 1)

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