Question

A beam of light passes through a sheet of plastic and out into the air. The beam makes ${\theta }_1$ angles with the normal as it passes through the plastic and the angle the beam of light makes with the normal as it passes into the air is ${\theta }_2$. The refractive index for air is $1$ and The refractive index for plastic is $2$.
Find out the minimum incident angle for which the light will undergo total internal reflection in the plastic?

• A. ${\sin }^{-1}\frac{1}{4}$
• B. ${\sin }^{-1}\frac{1}{2}$
• C. ${\sin }^{-1}2$
• D. $0{\hat{A}}^o$
• E. $90{\hat{A}}^o$

Answer 1

Answer: (B)

${\sin }^{-1}\frac{1}{2}$

Given - refractive index of first (denser) medium ${\mu }_1=2$ , refractive index of second (rarer) medium ${\mu }_2=1$

for total internal reflection,

or $sinC={\mu }_2/{\mu }_1=1/2$

or $C={\sin }^{-1}\frac{1}{2}$

where $C$ is the minimum angle of incidence for total internal reflection , called critical angle.