Physics · Geometrical Optics
JEE Advanced 2023 — Paper 1 — Question 18
A plane polarized blue light ray is incident on a prism such that there is no reflection from the surface of the prism. The angle of deviation of the emergent ray is (see Figure-1). The angle of minimum deviation for red light from the same prism is (see Figure-2). The refractive index of the prism material for blue light is . Which of the following statement(s) is(are) correct?
- Option A:Correct
The blue light is polarized in the plane of incidence
- Option B:Correct
The angle of the prism is .
- Option C:
The refractive index of the material of the prism for red light is .
- Option D:Correct
The angle of refraction for blue light in air at the exit plane of the prism is
Answer: A, B, D
Step-by-step solution
For Figure -1 (Blue light)
\delta=\mathrm{i}+\mathrm{e}-\mathrm{A} \Rightarrow 60^{\circ}=60^{\circ}+\mathrm{e}-\mathrm{A} \Rightarrow \mathrm{e}=\mathrm{A} …(i) \end{gathered}$$ At incident surface, $\sin 60^{\circ}=\sqrt{3} \operatorname{sinr}_{1}$ $\Rightarrow \mathrm{r}_{1}=30^{0}$ $\because \mathrm{r}_{1}+\mathrm{r}_{2}=\mathrm{A}$ $\Rightarrow \mathrm{r}_{2}=\mathrm{A}-30^{\circ}$ At emergent surface, $\sqrt{3} \sin \left(A-30^{\circ}\right)=\sin A$ $\frac{3}{2} \sin \mathrm{~A}-\frac{\sqrt{3}}{2} \cos \mathrm{~A}=\sin \mathrm{A}$ $\Rightarrow \tan A=\sqrt{3}$ $\mathrm{A}=60^{\circ}$ $\Rightarrow \mathrm{e}=60^{\circ}$ For Figure 2 (red line) For minimum deviation $\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \frac{A}{2}}=\mu_{R}$ $\Rightarrow \frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\mu_{\mathrm{R}}$ Or, $\mu_{\mathrm{R}}=\sqrt{2}$Answer key and solution verified before publishing.
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- Exam
- JEE Advanced 2023
- Paper
- Paper 1
- Subject
- Physics
- Chapter
- Geometrical Optics
- Topic
- Apparent Depth, Glass Slab, Prism and Dispersion