Question 47: Ray Optics

Question

A light ray is incident on the surface of a sphere of refractive index $n$ at an angle of incidence $	heta_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
List-II

(P) If $n = 2$ and $\alpha = 180^\circ$, then all the possible values of $	heta_0$ will be
(1) $30^\circ$ and $0^\circ$

(Q) If $n = \sqrt{3}$ and $\alpha = 180^\circ$, then all the possible values of $	heta_0$ will be
(2) $60^\circ$ and $0^\circ$

(R) If $n = \sqrt{3}$ and $\alpha = 180^\circ$, then all the possible values of $\phi_0$ will be
(3) $45^\circ$ and $0^\circ$

(S) If $n = \sqrt{2}$ and $	heta_0 = 45^\circ$, then all the possible values of $\alpha$ will be
(4) $150^\circ$

(5) $0^\circ$

Accepted Answer

$$\begin{aligned}
&\begin{aligned}
& \alpha=\left(	heta_0-\phi_0ight)+\left(180-2 \phi_0ight)+\left(	heta_0-2 \phi_0ight) \
& \alpha=180+2 	heta_0-4 \phi_0
\end{aligned}\
&\begin{aligned}
& 	ext { (P) } \alpha=180+2 	heta_0-4 \phi_0 \
& 180=180+2 	heta_0-4 \phi_0 \Rightarrow 	heta_0=2 \phi_0 \quad 	ext{... (i}\
& \sin 	heta_0=2 \sin \phi_0 \quad 	ext{... (ii}
\end{aligned}
\end{aligned}$$
From (i) & (ii)
$$\begin{aligned}
& \sin 	heta_0=2 \sin \left(	heta_0 / 2ight) \Rightarrow \cos \left(\frac{	heta_0}{2}ight)=1 \
& \frac{	heta_0}{2}=0 \
& \Rightarrow 	heta_0=0
\end{aligned}$$
$$\begin{aligned}
& 	ext { (Q) } 	heta_0=2 \phi_0 \quad 	ext{... (i)}\
& \sin 	heta_0=\sqrt{3} \sin \phi_0 \quad 	ext{... (ii)}
\end{aligned}$$
From (i) & (ii)
$$\begin{aligned}
& \sin 	heta_0=\sqrt{3} \sin \left(\frac{	heta_0}{2}ight) \
& \Rightarrow \cos \left(\frac{	heta_0}{2}ight)=\frac{\sqrt{3}}{2} \
& \frac{	heta_0}{2}=30,150 \
& 	heta_0=60,300 	ext { (Rejected) } \
& 	heta_0=60 \& 0
\end{aligned}$$
$$\begin{aligned}
& 	ext { (R) } 	heta_0=2 \phi_0 \
& \sin 	heta_0=\sqrt{3} \sin \phi_0 \
& \sin 2 	heta_0=\sqrt{3} \sin \phi_0 \
& \cos \phi_0=\frac{\sqrt{3}}{2} \
& \phi_0=30,150 	ext { (Rejected) }
\end{aligned}$$
$$\begin{aligned}
& \phi_0=30 \& 0 \quad 	ext{... (i)}\
& (S) \sin 45=\sqrt{2} \cos \phi_0 \
& \cos \phi_0=1 / 2 \
& \phi_0=60 \
& \alpha=180+2 	heta_0-4 \phi_0
\end{aligned}$$
$$\begin{aligned}
& \alpha=180+90-120  \quad 	ext{... (iv)}\
& =180-30 ; \alpha=150^{\circ}
\end{aligned}$$

Answer

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

Answer

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

Answer

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

Answer

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

Ray Optics

Choose the correct answer: