Question

A ray $R_1$ is incident on the plane surface of the glass slab (kept in air) of refractive index $\sqrt{2}$ at an angle of incidence equal to the critical angle for this air glass system. The refracted ray $R_2$ undergoes partial reflection and refraction at the other surface. The angle between reflected ray $R_3$ and the refracted ray $R_4$ at that surface is:

• A. ${45}^o$
• B. ${135}^o$
• C. ${105}^o$
• D. ${75}^o$

Answer 1

The correct option is C ${105}^o$

To find: Angle between $R_3$ & $R_4=?$

Angle of incident $=$ critical angle

Solution: Critical angle ${\theta }_c=\sin \left(\frac{1}{\mu }\right)=\sin \left(\frac{1}{\sqrt{2}}\right)$

$i={\theta }_c={45}^o$

Applying snell's law at $1$st surface

${\mu }_1\sin i={\mu }_2\sin r$

$1\times \sin 45=\sqrt{2}\sin r$

$\Rightarrow r={30}^{\circ }$

Applying snell's law at $2$nd surface

${\mu }_2\sin r={\mu }_1\sin e$

$\sqrt{2}\times \sin 30=1\times \sin e$

$\Rightarrow \sin e=\frac{1}{\sqrt{2}}$

$\Rightarrow e={45}^o$.

Angle between $R_3$ & $R_4=180-r-e$

$=180-30-45=105$.