Question

The figure given below shows a solid glass block of refractive index $1.5$. The lower surface of the block is silvered and behaves as a reflecting mirror. A white light ray $AB$ incidents on the glass block at an angle of incidence ${40}^{\circ }$. Copy the diagram and trace the path of reflected and refracted rays from the upper and lower surface of the block.

Answer 1

From the above diagram we can find all the unknown

If reflection occurs at surface PS then

$\angle i=\angle rr={40}^0$ reflection angle

By snell's law

$\mu =\frac{Sini}{Sinr}1.5=\frac{sin{40}^0}{sinr}$

$sinr=0.428\Rightarrow r={25.34}^0\simeq {26}^0$

second when refracted light from B goes to surface QR and reflects back and come back to surface PS here both refraction and reflection occurs.

thus,

$\frac{1}{\mu }=\frac{Sinr}{Sine}\frac{1}{1.5}=\frac{sin{26}^0}{sine}sine=0.642\Rightarrow e={40}^0$