QJEEAdv15Ph61d3: JEE Advanced 2015 Physics - Class 12 JEEAdv15Ph61d3 Solution: For two structures namely S 1 with n 1 = and n 2 = , and S 2…

Let the whole structure be placed in the medium of refractive index n₀. From geometry, if the angle of incidence at Q is $\theta$ then the angle of refraction at P is 90 $^\circ$ $-$ $\theta$ .

The ray will undergo total internal reflection at Q if the angle of incidence at Q is greater than or equal to the critical angle i.e.,

$\theta$ $\ge$ $\theta$ _c = sin ^{$-$ 1}(n₂ / n₁). ........ (1)

Apply Snell's law for refraction at P to get

n₀ sin i = n₁ sin(90 $^\circ$ $-$ $\theta$ ) = n₁ cos $\theta$

= ${n_1}\sqrt {1 - {{\sin }^2}\theta }$ . ...... (2)

From equation (2), the angle of incidence is maximum (i = i_m) when $\theta$ is minimum i.e., when $\theta$ = $\theta$ _c (from equation (1)). Thus, the numerical aperture is given by

$NA = \sin {i_m} = {{{n_1}\sqrt {1 - {{\sin }^2}{\theta _c}} } \over {{n_0}}} = {{\sqrt {n_1^2 - n_2^2} } \over {{n_0}}}$ ........ (3)

Substitute ${n_1} = \sqrt {45} /4$ and ${n_2} = 3/2$ in equation (3) to get the numerical aperture for the structure S₁ is

$N{A_1} = {{\sqrt {45/16 - 9/4} } \over {{n_0}}} = {3 \over {4{n_0}}}$ ...... (4)

Similarly, substitute n₁ = 8/4 and n₂ = 7/5 in equation (3) to get the numerical aperture for the structure S₂ as

$N{A_2} = {{\sqrt {64/25 - 49/25} } \over {{n_0}}} = {{\sqrt {15} } \over {5{n_0}}}$ ...... (5)

In case (A), substitute n₀ = 4/3 in equation (4) to get NA₁ = 9/16 and substitute n₀ = 16/3 $\sqrt15$ in equation (5) to get NA₂ = 9/16.

In case (B), substitute n₀ = 6/ $\sqrt15$ in equation (4) to get NA₁ = $\sqrt15$ /8 and substitute n₀ = 4/3 in equation (5) to get NA₂ = 3 $\sqrt15$ /20.

In case (C), substitute n₀ = 1 in equation (4) to get NA₁ = 3/4 and substitute n₀ = 4/ $\sqrt15$ equation (5) to get NA₂ = 3/4.

In case (D), substitute n₀ = 1 in equation (4) to get NA₁ = 3/4 and substitute n₀ = 4/3 equation (5) to get NA₂ = 3 $\sqrt15$ /20.