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Physics · Geometrical Optics

JEE Advanced 2021 — Paper 2 — Question 3

For a prism of prism angle θ=60∘\theta=60^{\circ}, the refractive indices of the left half and the right half are, respectively, n1\mathrm{n}_{1} and n2(n2≥n1)\mathrm{n}_{2}\left(\mathrm{n}_{2} \geq \mathrm{n}_{1}\right) as shown in the figure. The angle of incidence ii is chosen such that the incident light rays will have minimum deviation if n1=n2=n=1.5n_{1}=n_{2}=n=1.5. For the case of unequal refractive indices, n1=nn_{1}=n and n2=n_{2}= n+Δn\mathrm{n}+\Delta \mathrm{n} (where Δn≪n\Delta \mathrm{n} \ll \mathrm{n} ), the angle of emergence e=i+Δee=i+\Delta e. Which of the following statement(s) is(are) correct?

Question figure
  1. Option A:

    The value of Δe\Delta e (in radians) is greater than that of Δn\Delta \mathrm{n}

  2. Option B:

    Δe\Delta e is proportional to Δn\Delta \mathrm{n}

    Correct
  3. Option C:

    Δe\Delta e lies between 2.0 and 3.0 milliradians, if Δn=2.8×10−3\Delta \mathrm{n}=2.8 \times 10^{-3}

    Correct
  4. Option D:

    Δe\Delta e lies between 1.0 and 1.6 milliradians, if Δn=2.8×10−3\Delta \mathrm{n}=2.8 \times 10^{-3}

Answer: B, C

Step-by-step solution

Diagram at minimum deviation for n1=n2=n\mathrm{n}_{1}=\mathrm{n}_{2}=\mathrm{n} n=1.5\mathrm{n}=1.5 r1=r2=θ/2=30∘r_{1}=r_{2}=\theta / 2=30^{\circ}

for face AQ nsinr⁡2=sin⁡en \operatorname{sinr}_{2}=\sin \mathrm{e} 1.5sin⁡30∘=32×12=sin⁡e1.5 \sin 30^{\circ}=\frac{3}{2} \times \frac{1}{2}=\sin \mathrm{e}

sin⁡e=34,cos⁡e=74\sin \mathrm{e}=\frac{3}{4}, \quad \cos \mathrm{e}=\frac{\sqrt{7}}{4} When n2n_{2} is given small variation there will be no change in path of light ray inside prism.

As deviation on face AC is zero. So, r2=30∘r_{2}=30^{\circ} Now for face AQ n2sin⁡30∘=sin⁡e\mathrm{n}_{2} \sin 30^{\circ}=\sin \mathrm{e} for small change in n2n_{2}

change in ee is given by dn2sin⁡30∘=cos⁡ede\mathrm{dn}_{2} \sin 30^{\circ}=\cos \mathrm{e} \mathrm{de} or dn2=Δn\mathrm{dn}_{2}=\Delta \mathrm{n} \quad

de =Δe=\Delta \mathrm{e} Δnsin⁡30∘=cos⁡eΔe\Delta \mathrm{n} \sin 30^{\circ}=\cos \mathrm{e} \Delta \mathrm{e} Δn12=74Δe\Delta n \frac{1}{2}=\frac{\sqrt{7}}{4} \Delta e

Δn=72Δe\Delta \mathrm{n}=\frac{\sqrt{7}}{2} \Delta \mathrm{e} ...(i) Δn>Δe\Delta n>\Delta e Δn∝Δe\Delta n \propto \Delta e

Hence, option (B) is correct. Δe=2.8×10−3×27\Delta \mathrm{e}=\frac{2.8 \times 10^{-3} \times 2}{\sqrt{7}} Hence, option (C) is correct.

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2021
Paper
Paper 2
Subject
Physics
Chapter
Geometrical Optics
Topic
Apparent Depth, Glass Slab, Prism and Dispersion