A light ray is incident on a glass slab of thickness 4 √3 cm and refractive index √2 The angle of incidence is equal to the critical angle for the glass slab with air. The lateral displacement of ray after passing through glass slab is ______ cm.
(Given sin 15°=0.25)

Solution

Physics sir · 4 steps

Critical angle gives i = 45°, Snell's law gives r = 30°, then use the slab-shift formula.

1. Angle of incidence. For glass to air, sin C = 1/μ = 1/√2, so C = 45°. The ray enters the slab at i = 45°.
2. Angle of refraction. Snell's law at the top face gives sin r = (sin 45°)/√2 = 1/2, so r = 30°. That makes i - r = 15°.

   d = (t sin(i - r))/(cos r)

3. Put in numbers. d = (4√3 × 0.25)/(√3/2) = 4√3 × 0.25 × 2/√3. The √3 cancels, leaving 4 × 0.25 × 2.
4. So. The lateral displacement is 2 cm.

The common slip: 1.73. Leaving out the cos r in the denominator gives 4√3 × 0.25 ≈ 1.73 instead of 2.

The idea to keep. The ray travels a distance t/cos r inside the slab. The sideways shift is that path times sin(i - r). Find i and r first and the rest is substitution.

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