Question

A rectangular glass slab ABCD of refractive index $n_1$, is immersed in water of refractive index $n_2$($n_1>n_2$). A ray of light is incident at the surface AB of the slab as shown in the figure. The maximum value of angle of incidence ${\alpha }_{max}$, such that the ray comes out only from the other surface CD, is given by

• A. ${sin}^{-1}\left[\frac{n_1}{n_2}cos\left({sin}^{-1}\frac{n_2}{n_1}\right)\right]$
• B. ${sin}^{-1}\left[n_{1}cos\left({sin}^{-1}\frac{1}{n_2}\right)\right]$
• C. ${sin}^{-1}\left(\frac{n_1}{n_2}\right)$
• D. ${sin}^{-1}\left(\frac{n_2}{n_1}\right)$

Answer 1

The correct option is A ${sin}^{-1}\left[\frac{n_1}{n_2}cos\left({sin}^{-1}\frac{n_2}{n_1}\right)\right]$

For the max value of $\alpha$ such that the ray should should come out from CD is it should suffer limiting of total internal reflection at the interface AD . so the ray graces at the interface AD

angle of incidence at interface AD is $si{n}^{-1}\frac{1}{{\mu }_r}=si{n}^{-1}\frac{n_2}{n_1}$

apply snell's law at boundary AB
${\mu }_{1}sin{\alpha }_1={\mu }_{2}sin{\alpha }_2$
$n_{2}sin\alpha =n_{1}sin(90-critica{l}_{a}ngle)$

$sin\alpha =\frac{n_1}{n_2}cos\left(si{n}^{-1}\frac{n_2}{n_1}\right)$

$\alpha =si{n}^{-1}\left(\frac{n_1}{n_2}cos\right(si{n}^{-1}\frac{n_2}{n_1}\left)\right)$.