QJEEAdv16Phca6d: JEE Advanced 2016 Physics - Class 12 JEEAdv16Phca6d Solution: A parallel beam of light is incident from air at an anglea on…

Applying Snell's law at M,

$n = {{\sin \alpha } \over {\sin {r_1}}} \Rightarrow \sqrt 2 = {{\sin 45^\circ } \over {\sin {r_1}}}$

$\Rightarrow \sin {r_1} = {{\sin 45^\circ } \over {\sqrt 2 }} = {{1/\sqrt 2 } \over {\sqrt 2 }} = {1 \over 2}$

${r_1} = 30^\circ$

$\sin {\theta _c} = {1 \over n} = {1 \over {\sqrt 2 }} \Rightarrow {\theta _c} = 45^\circ$

Let us take ${r_2} = {\theta _c} = 45^\circ$ for just satisfying the condition of TIR.

In $\Delta PNM$ ,

$\theta + 90 + {r_1} + 90 - {r_2} = 180^\circ$

or $\theta = {r_2} - {r_1} = 45^\circ - 30^\circ = 15^\circ$

Note If $\alpha$ = 45 $^\circ$ (the given value). Then, r₁ > 30 $^\circ$ (the obtained value)

{r_2} > {\theta _c} (as r₂ $-$ r₁ = $\theta$ or r₂ = $\theta$ + r₁)

or TIR will take place. So, for taking TIR under all conditions $\alpha$ should be greater than 45 $^\circ$ or this is the minimum value of $\alpha$ .