Question

Light is incident from glass $(\mu =1.5)$ to water $(\mu =1.33)$. Find the range of the angle of deviation for which there are two angles of incidence.

Answer 1

${\mu }_g=1.5=\frac{3}{2},{\mu }_w=1.33=\frac{4}{3}$

For two angles of incidence,

1. When light passes straight, through Normal,

$\Rightarrow$ Angle of incidence $=0^{c}irc$

$\Rightarrow$ Angle of refraction $=0^{c}irc$

$\Rightarrow$ Angle of deviation $=0^{c}irc$

2. When light is incident at critical angle,

$\frac{sinC}{sinr}=\frac{{\mu }_a}{{\mu }_g}$

(since light passing from glass to water)

$\Rightarrow sinC=\frac{8}{9}$

$\Rightarrow C=si{n}^{-1}\left(\frac{8}{9}\right)={62.73}^{\circ }$

$\Rightarrow$ Angle of deviation

$={90}^{\circ }-C$

$=90-si{n}^{-1}\frac{8}{9}$

$=co{s}^{-1}\frac{8}{9}$

$={37.27}^{\circ }$

Here, if the angle of increased beyond critical angle, total internal reflection occurs and deviation decreases. So, the range of deviation is 0 to $co{s}^{-1}\left(\frac{8}{9}\right)$