Question

A ray of light travels from air into a perspex block of refractive index 1.5 at an angle of incidence of ${45}^0$. Calculate the angle of the refraction of the ray at air-to perspex boundary and at perpex-to-air boundary.

• A. ${28}^0{7}^{\prime }$
• B. ${25}^0{87}^{\prime }$
• C. ${25}^0{7}^{\prime }$
• D. ${28}^0{27}^{\prime }$

Answer 1

The correct option is A ${28}^0{7}^{\prime }$

${\mu }_{perspex}=1.5=\frac{sin{x}^0}{sinr}$

and $\angle i={45}^0$

$\therefore \frac{sin{45}^0}{sinr}=1.5$

$sinr=\frac{1}{\sqrt{2}\times 1.5}$

$r={28}^0{7}^{\prime }$

The angle of refraction of the ray at air to perpex is ${28}^0{7}^{\prime }$.

The angle of refraction at perpex to air boundary is angle of incident at air to perpex boundary i.e., ${45}^0.$