Question

the critical angle for a medium is 45' the refractive index of the medium is

Answer 1

Answer:

$\textcolor{red}{\sqrt{2}}$

Explanation:

$\mathrm{\theta _c\: be \: the \: critical \: angle,}\\</p><p>\sin \theta= \frac{1}{n}\: (n=refractive \: index)\\</p><p>\implies \frac{1}{n}=\sqrt2\\</p><p>\implies n=\sqrt 2$