Verified Solution • 2025–26 Session moderate

QJEEAdv24Ph7340: JEE Advanced 2024 Physics - Class 12 JEEAdv24Ph7340 Solution: A light ray is incident on the surface of a sphere of…

A light ray is incident on the surface of a sphere of refractive index nn at an angle of incidence θ0\theta_0 . The ray partially refracts into the sphere with angle of refraction ϕ0\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 α\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=2n = 2 and α=180∘\alpha = 180^\circ , then all the possible values of θ0\theta_0 will be (1) 30∘30^\circ and 0∘0^\circ
(Q) If n=3n = \sqrt{3} and α=180∘\alpha = 180^\circ , then all the possible values of θ0\theta_0 will be (2) 60∘60^\circ and 0∘0^\circ
(R) If n=3n = \sqrt{3} and α=180∘\alpha = 180^\circ , then all the possible values of ϕ0\phi_0 will be (3) 45∘45^\circ and 0∘0^\circ
(S) If n=2n = \sqrt{2} and θ0=45∘\theta_0 = 45^\circ , then all the possible values of α\alpha will be (4) 150∘150^\circ
(5) 0∘0^\circ

Multiple Choice Options

  • A P→5;Q→2;R→1;S→4\mathrm{P} \rightarrow 5 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 4
  • B P→5;Q→1;R→2;S→4\mathrm{P} \rightarrow 5 ; \mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 4
  • C P→3;Q→2;R→1;S→4\mathrm{P} \rightarrow 3 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 4
  • D P→3;Q→1;R→2;S→5\mathrm{P} \rightarrow 3 ; \mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 5

Step-by-Step Verified Solution

JEE Advanced 2024 Paper 1 Online Physics - Geometrical Optics Question 2 English Explanation

amp;amp;α=(θ0−ϕ0)+(180−2ϕ0)+(θ0−2ϕ0)amp;α=180+2θ0−4ϕ0amp;amp; (P) α=180+2θ0−4ϕ0amp;180=180+2θ0−4ϕ0⇒θ0=2ϕ0... (iamp;sin⁡θ0=2sin⁡ϕ0... (ii\begin{aligned}&\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}\\&\begin{aligned}& \text { (P) } \alpha=180+2 \theta_0-4 \phi_0 \\& 180=180+2 \theta_0-4 \phi_0 \Rightarrow \theta_0=2 \phi_0 \quad \text{... (i}\\& \sin \theta_0=2 \sin \phi_0 \quad \text{... (ii}\end{aligned}\end{aligned}

From (i) & (ii)

amp;sin⁡θ0=2sin⁡(θ0/2)⇒cos⁡(θ02)=1amp;θ02=0amp;⇒θ0=0\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}

amp; (Q) θ0=2ϕ0... (i)amp;sin⁡θ0=3sin⁡ϕ0... (ii)\begin{aligned}& \text { (Q) } \theta_0=2 \phi_0 \quad \text{... (i)}\\& \sin \theta_0=\sqrt{3} \sin \phi_0 \quad \text{... (ii)}\end{aligned}

From (i) & (ii)

amp;sin⁡θ0=3sin⁡(θ02)amp;⇒cos⁡(θ02)=32amp;θ02=30,150amp;θ0=60,300 (Rejected) amp;θ0=60&0\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}

amp; (R) θ0=2ϕ0amp;sin⁡θ0=3sin⁡ϕ0amp;sin⁡2θ0=3sin⁡ϕ0amp;cos⁡ϕ0=32amp;ϕ0=30,150 (Rejected) \begin{aligned}& \text { (R) } \theta_0=2 \phi_0 \\& \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}

amp;ϕ0=30&0... (i)amp;(S)sin⁡45=2cos⁡ϕ0amp;cos⁡ϕ0=1/2amp;ϕ0=60amp;α=180+2θ0−4ϕ0\begin{aligned}& \phi_0=30 \& 0 \quad \text{... (i)}\\& (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}

amp;α=180+90−120... (iv)amp;=180−30;α=150∘\begin{aligned}& \alpha=180+90-120 \quad \text{... (iv)}\\& =180-30 ; \alpha=150^{\circ}\end{aligned}

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