Question

A light beam is travelling from Region I to Region IV as shown in the figure below: The refractive indices of Regions I, II, III and IV are `n_0, n_0/2,n_0/6` and `n_0/8` respectively. The angle of incidence θ for which the beam just misses entering Region IV is:

Options

• `sin^-1 (3/4)`

• `sin^-1 (1/8)`

• `sin^-1 (1/4)`

• `sin^-1 (1/3)`

Answer 1

`bb(sin^-1 (1/8))`

Explanation:

Refractive indices:

Region I: n₀

Region II: `n_0/2`

Region III: `n_0/6`

Region IV: `n_0/8`

sin θ_c = `(n_0/8)/(n_0/6) = 3/4`

Step 1: Critical angle at the region III–IV interface:

For the light ray to just enter region IV, it must strike the region III–IV boundary at the critical angle θ_c​:

`sin θ_c = n_4/n_3`

= `(n_0/8)/(n_0/6)`

= `6/8`

= `3/4`

Step 2: Apply Snell’s law from region I to III directly:

From region I to III (combining both interfaces), Snell’s law:

`n_0 sin θ = n_0/6 sin θ_c`

`sin θ = 1/6 xx sin θ_c`

= `1/6 xx 3/4`

= `1/8`

θ = `sin^-1 (1/8)`