Question

(a) Calculate the critical angle for glass air interface if a ray of light incident on a glass surface is deviated through ${15}^o$ when angle of incident is ${45}^o$.

(b) A ray of light travels from denser medium having refractive index $\sqrt{2}$ to air, What should be the angle of refracted when angle of incident is ${30}^0$.

Answer 1

(a).

Given that,

Incident angle $i={45}^0$

Let critical angle be $i_c$

Now,

$\sin i_c=\mu$

Where, $\mu$ is the refractive index of glass

Now, again

$r={45}^0-{15}^0$

$r={30}^0$

$\frac{\sin i}{\sin r}=\mu$

$\mu =\frac{\sin {45}^0}{\sin {30}^0}$

$\mu =\sqrt{2}$

$\mu =1.4$

Now, the critical angle is

$\sin i_c=\frac{1}{\mu }$

${\mathrm{sini}}_c=\frac{1}{1.4}$

$i_c={\sin }^{-1}\left(0.71\right)$

$i_c={45.2}^0$

Hence, the critical angle is ${45.2}^0$

(b). given that,

Incident angle $i={45}^0$

Now, from Snell’s formula

$\frac{\sin i}{\sin r}=\mu$

$\frac{\sin 45}{\sin r}=\sqrt{2}$

$\frac{1}{\sqrt{2}\times \sqrt{2}}=\sin r$

$\sin r=\frac{1}{2}$

$r={30}^0$

Hence, the refractive angle is ${30}^0$