# Question #354d2

Jan 14, 2018

41.81 degrees

#### Explanation: Critical angle of light is that one,if at that angle light falls at an interface,while going from a solid to lighter media,it will not enter the media next to the interface,rather go along the interface,

So,using Snell's law we can say,
$u 1 \sin \theta$ = $u 2 \sin \alpha$
Given, $u 2$ = $\left(\frac{3}{2}\right)$
And $\sin \theta$ = 1
So, $\sin \alpha$ =$\frac{2}{3}$ or $\alpha$ = 41.81 degrees

Jan 14, 2018

the critical angle is ${i}_{c} = 42.067$

#### Explanation:

critical angle ${i}_{c}$ is given as
$\sin \left({i}_{c}\right) = \frac{1}{n} _ 21$ where ${n}_{21} = {n}_{2} / {n}_{1}$
$n$ is the refractive index of that particular media
$h e r e {n}_{2} = \frac{3}{2} \mathmr{and} {n}_{1} = 1$ substituting in the formula we get
$\sin \left({i}_{c}\right) = \left(\frac{1}{\frac{3}{2}}\right) = \left(\frac{2}{3}\right) \Rightarrow {i}_{c} = {\sin}^{-} 1 \left(\frac{2}{3}\right) = {\sin}^{-} 1 \left(0.67\right)$
$\Rightarrow {i}_{c} = 42.067$