Question

A rectangular glass slab ABCD of refractive index ${\mu }_1$ is immersed in water of refractive index ${\mu }_2({\mu }_2<{\mu }_1)$. A ray of light is incident at the surface AB of the slab as shown. The maximum value of the angle of incidence ${\alpha }_{max}$ such that the ray comes out from the other surface CD is given by

• A. $si{n}^{-1}\left[\frac{{\mu }_1}{{\mu }_2}cos\left(si{n}^{-1}\left(\frac{{\mu }_2}{{\mu }_1}\right)\right)\right]$
• B. $si{n}^{-1}\left[{\mu }_{1}cos\left(si{n}^{-1}\left(\frac{1}{{\mu }_2}\right)\right)\right]$
• 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 $si{n}^{-1}\left[\frac{{\mu }_1}{{\mu }_2}cos\left(si{n}^{-1}\left(\frac{{\mu }_2}{{\mu }_1}\right)\right)\right]$

For the ray to emerge from the face CD, it at first must undergo total internal reflection at face AD i-e, $r_1>{\theta }_c$

$r_1=si{n}^{-1}(\frac{{\mu }_2}{{\mu }_1})$ .........................(1)

Also from geometry, $r=90-r_1⟹sinr=cos{r}_1$ .............(2)

Snell's law for face AB, ${\mu }_2\sin {\alpha }_{max}={\mu }_{1}sinr$

Thus from (1) & (2), ${\mu }_{2}sin{\alpha }_{max}={\mu }_{1}cos(si{n}^{-1}(\frac{{\mu }_2}{{\mu }_1}))$

Thus ${\alpha }_{max}=si{n}^{-1}[\frac{{\mu }_1}{{\mu }_2}cos(si{n}^{-1}(\frac{{\mu }_2}{{\mu }_1}))]$