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Physics · Geometrical Optics

JEE Main 2026 — 24 January, Morning Shift — Question 29

The exit surface of a prism with refractive index nn is coated with a material having refractive index n2\frac{\mathrm{n}}{2}. When this prism is set for minimum angle of deviation it exactly meets the condition of critical angle. The prism angle is ____\_\_\_\_

  1. Option A:

    60∘60^{\circ}

    Correct
  2. Option B:

    15∘15^{\circ}

  3. Option C:

    30∘30^{\circ}

  4. Option D:

    45∘45^{\circ}

Answer: A

Step-by-step solution

i=e&r=A/2\mathrm{i}=\mathrm{e} \& \mathrm{r}=\mathrm{A} / 2 for minimum deviation Sin⁡r=Sin⁡θC\operatorname{Sin} \mathrm{r}=\operatorname{Sin} \theta_{\mathrm{C}} Sin⁡r=n/2n\operatorname{Sin} \mathrm{r}=\frac{\mathrm{n} / 2}{\mathrm{n}} Sin⁡r=12\operatorname{Sin} \mathrm{r}=\frac{1}{2} Sin⁡A2=sin⁡30∘\operatorname{Sin} \frac{\mathrm{A}}{2}=\sin 30^{\circ} A2=30∘\frac{\mathrm{A}}{2}=30^{\circ} A=60∘\mathrm{A}=60^{\circ}

figure

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
24 January, Morning Shift
Subject
Physics
Chapter
Geometrical Optics
Topic
Apparent Depth, Glass Slab, Prism and Dispersion

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