Answer:
Option D
Explanation:
for e → i
$45^{0}>\theta_{c}$
$\sin 45^{0}>\sin\theta_{c}$
$\frac{1}{\sqrt{2}}>\frac{\mu_{2}}{\mu_{1}}$
$\mu_{1}> \sqrt{2\mu_{2}}$
For e $\rightarrow$ f
angle of refraction is lesser than angle of incidence , so $\mu_{2}> \mu_{1}$ and $\mu_{2}> \mu_{3}$
for e $\rightarrow$ g $\mu_{1}= \mu_{2}$
for e $\rightarrow$ h , $\mu_{2}< \mu_{1}$ < $\sqrt{2}\mu_{2}$ and $\mu_{2} >\mu_{3} $