CBSE Class 12 Physics 2025: question 19

Hints

Open one hint at a time and try again before the next.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. The ray travels along the curved surface, so it meets that surface at exactly the critical angle $C$: $\sin C=\frac{1}{n}=\frac{\sqrt{3}}{2}$, so $C=60^{\circ }$.

   ½ mark

2. The end face is perpendicular to the curved surface, so the angle of refraction at the face is $r=90^{\circ }-C=30^{\circ }$.

   ½ mark

3. Snell's law at the air–rod face: $1\times \sin \theta =n\sin r=\frac{2}{\sqrt{3}}\times \sin 30^{\circ }$.

   ½ mark

4. $\sin \theta =\frac{2}{\sqrt{3}}\times \frac{1}{2}=\frac{1}{\sqrt{3}}$, so $\theta ={\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)\approx 35.3^{\circ }$.

   ½ mark

Answer: θ = sin⁻¹(1/√3) ≈ 35.3°

Revise the chapter

• Chapter: Ray Optics and Optical Instruments (notes, formulas, where marks are lost and every past question)
• Topic: Total Internal Reflection