Question

A rectangular slab $ABCD$ of refractive index ${\mu }_1$is immersed in a liquid of refractive index ${\mu }_2$ (${\mu }_2<{\mu }_1)$. A ray of light is incident on the surface $AB$ of the slab as shown in the figure.The maximum value of incidence such that the ray comes out only from the surface $CD$, is given by

• A. $si{n}^{-1}\left(\frac{\sqrt{{\mu }_1^2-{\mu }_2^2}}{{\mu }_2}\right)$
• B. $si{n}^{-1}\left(\frac{\sqrt{{\mu }_1^2-{\mu }_2^2}}{{\mu }_1}\right)$
• C. $si{n}^{-1}\left(\frac{{\mu }_1}{{\mu }_2}\right)$
• D. $si{n}^{-1}\left(\frac{{\mu }_2}{{\mu }_1}\right)$

Answer 1

The correct option is A $si{n}^{-1}\left(\frac{\sqrt{{\mu }_1^2-{\mu }_2^2}}{{\mu }_2}\right)$

For the ray to come out of the face CD, it should undergo total internal reflection at interface AD.
At the first interface using Snell's law we have
${\mu }_{2}sini={\mu }_{1}sinr$
For the second interface for total internal reflection we have
${\mu }_{1}sin{r}^{\prime }=\mu sin{90}^o$
or
${\mu }_1\sqrt{1-si{n}^{2}r}={\mu }_2$
or
${\mu }_1\sqrt{1-(\frac{{\mu }_2}{{\mu }_1}{)}^{2}si{n}^{2}i}={\mu }_2$
or
$1-(\frac{{\mu }_2}{{\mu }_1}{)}^{2}si{n}^{2}i=(\frac{{\mu }_2}{{\mu }_1}{)}^2$
or
$1+si{n}^{2}i=(\frac{{\mu }_1}{{\mu }_2}{)}^2$
or
$si{n}^{2}i=(\frac{{\mu }_1}{{\mu }_2}{)}^2-1$
or
$i=si{n}^{-1}\frac{\sqrt{{\mu }_1^2-{\mu }_2^2}}{{\mu }_2}$