A ray of laser of a wavelength 630 nm is incident at an angle of 30° at the diamond-air interface. It is going from diamond to air. The refractive index of diamond is 2.42 and that of air is 1. Choose the correct option.

Solution

Physics sir · 4 steps

Before you find an angle of refraction, check that sin r does not come out bigger than 1.

1. Apply Snell's law. The ray goes from diamond (n₁ = 2.42) into air (n₂ = 1) at i = 30°. The wavelength of 630 nm does not change anything here.

   sin r = (n₁ sin i)/n₂ = (2.42 × 0.5)/1 = 1.21

2. Impossible sine. The sine of an angle can never exceed 1, so no refracted ray exists. The light is totally reflected back into the diamond.
3. Check critical angle. For diamond, sin C = 1/2.42 = 0.413, so C ≈ 24.4°. The angle of incidence, 30°, is larger than C, which confirms total internal reflection.
4. So. Refraction is not possible, so the answer is (c).

The idea to keep. When light goes from a denser to a rarer medium, compare i with the critical angle sin C = 1/n first. If i > C, there is no refracted ray at all, only total internal reflection.

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