Cross-section view of a prism is the equilateral triangle ABC in the figure. The minimum deviation is observed using this prism when the angle of incidence is equal to the prism angle. The time taken by light to travel from P (midpoint of BC) to A is ______________ × 10⁻¹⁰ s. (Given, speed of light in vacuum = 3 × 10⁸ m/s and cos30° = √3/2)

Solution

Physics sir · 3 steps

Minimum deviation fixes r = 30°, which gives μ. Then it is just distance over speed inside glass.

1. Refraction angle. The prism is equilateral, so A = 60°. At minimum deviation the ray passes symmetrically, so the refraction angle at each face is r = A/2 = 30°. The question gives i = A = 60°.

   μ = (sin i)/(sin r) = (sin 60°)/(sin 30°) = (√3/2)/(1/2) = √3

2. Speed and distance. Inside the prism v = c/μ = (3 × 10⁸)/√3 = √3 × 10⁸ m s⁻¹. The path PA is the altitude of the triangle, 10 cos 30° = 5√3 cm = 5√3 × 10⁻² m.
3. So. t = (5√3 × 10⁻²)/(√3 × 10⁸) = 5 × 10⁻¹⁰ s, so the answer is 5.

The idea to keep. At minimum deviation, always set r = A/2 and get μ from sin i / sin r. Light inside the medium travels at c/μ, not c.

Answer from the NTA final answer key. Solution by PrepWiser.