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WBJEEA light ray is incident on the surface of a sphere of refractive...
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.
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)
\(=180-30 ; \alpha=150^{\circ}\)
\(\begin{aligned} & \alpha=\left(\theta_0-\phi_0\right)+\left(180-2 \phi_0\right)+\left(\theta_0-2 \phi_0\right) \\ & \alpha=180+2 \theta_0-4 \phi_0\end{aligned}\)
\(\textbf{(P) } \alpha=180+2 \theta_0-4 \phi_0\)
\(180=180+2 \theta_0-4 \phi_0 \Rightarrow \theta_0=2 \phi_0\) ...(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)
\(=180-30 ; \alpha=150^{\circ}\)
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