Question

A transparent cube of side, made of a material of refractive index n₂, is immersed in a liquid of refractive index n₁ (n₁ < n₂). A ray is incident on the face AB at an angle θ (shown in the figure). Total internal reflection takes place at point E on the face BC. Then θ must satisfy:

Options

• `θ > sin^-1 sqrt((n_2^2)/(n_1^2) - 1)`

• `θ < sin^-1 sqrt((n_2^2)/(n_1^2) - 1)`

• `θ < sin^-1  n_1/n_2`

• `θ > sin^-1  n_1/n_2`

Answer 1

`bb(θ < sin^-1 sqrt((n_2^2)/(n_1^2) - 1))`

Explanation:

Let the angle of refraction for i = θ be α.

Then by Snell’s law

n₁ sin θ = n₂ sin α  ...(i)

For TIR at E

n₂ sin (90° − α) > n₁

From the equation (i),

sin α = `n_1/n_2` sin θ

cos α = `sqrt((n_2^2 - n_1^2 sin^2 θ)/n_2^2)`

= `sqrt((n_2^2)/(n_2^2) - (n_1^2 sin^2 θ)/n_2^2)`

= `sqrt(1 - n_1^2/n_2^2 sin^2 θ)`  ...(ii)

But sin (90° − α) = `cos α > n_1/n_2`

Hence from (ii),

`1 - n_1^2/n_2^2 sin^2 θ > n_1^2/n_2^2`

Or `sin^2 θ < (n_2^2/n_1^2 - 1)`

Or `θ < sin^-1 sqrt((n_2^2)/n_1^2 - 1)`