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Physics · Geometrical Optics

JEE Main 2026 — 23 January, Evening Shift — Question 37

A prism of angle 75∘75^{\circ} and refractive index 3\sqrt{3} is coated with thin film of refractive index 1.5 only at the back exit surface. To have total internal reflection at the back exit surface the incident angle can not be ____\_\_\_\_ . ( sin⁡15∘=0.25\sin 15^{\circ}=0.25 and sin⁡25∘=0.43\sin 25^{\circ}=0.43 )

  1. Option A:

    between 15∘15^{\circ} and 20∘20^{\circ}

  2. Option B:

    15∘15^{\circ}

  3. Option C:

    >25∘>25^{\circ}

    Correct
  4. Option D:

    <15∘<15^{\circ}

Answer: C

Step-by-step solution

r1+r2=75∘\mathrm{r}_{1}+\mathrm{r}_{2}=75^{\circ} For TIR at back surface 3sin⁡r2=32sin⁡90∘\sqrt{3} \sin \mathrm{r}_{2}=\frac{3}{2} \sin 90^{\circ} r2≥60∘\mathrm{r}_{2} \geq 60^{\circ} r1≤15∘\mathrm{r}_{1} \leq 15^{\circ} 1sin⁡i=3sin⁡15∘1 \sin \mathrm{i}=\sqrt{3} \sin 15^{\circ} sin⁡i=1.73×.25\sin \mathrm{i}=1.73 \times .25 sin⁡i=.433\sin \mathrm{i}=.433 i=25∘⇒i<25∘\mathrm{i}=25^{\circ} \Rightarrow \mathrm{i}<25^{\circ}

Solution figure

Answer key and solution verified before publishing.

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

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