A ray of light incident on the horizontal surface of a glass slab at 70°  just grazes the adjacent vertical surface after refraction,Find critical angle and refractive index of glass
 

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Please find below the solution to the asked query:

$$\begin{gathered} \frac{\sin i}{\sin r}=n \\ \frac{\sin 70}{\sin r}=n \\ \sin r=\frac{\sin 70}{n}=\frac{0.94}{n} \\ \text{In the next vertical surface the incident angle will be} {\theta }_i=90-r, \text{and } \\ \text{refractive angle is 90 degree. So we have} \\ \frac{\sin (90-r)}{\sin 90}=\frac{1}{n} \\ \cos r=\frac{1}{n} \\ \text{Now squaring and adding we get} \\ \frac{1}{n^2}+\frac{0.{94}^2}{n^2}=1 \\ \frac{1}{n^2}\left(1+0.{94}^2\right)=\frac{1.88}{n^2}=1 \\ n=\sqrt{1.88}=1.37 \\ \text{So critical angle is } \\ {\theta }_c={\sin }^{-1}\left(\frac{1}{n}\right)=46.88 \end{gathered}$$

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