Question

At what angle in degrees should a ray of light be incident on the face of a prism of refracting angle ${60}^o$ so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is $1.524$. Given $si{n}^{-1}\left(0.66\right)={41}^o$, $\sin {19}^o=0.3256$, ${\sin }^{-1}\left(0.4962\right)={30}^o$, $\frac{1}{1.5424}=0.66$

Answer 1

The refracted ray QR will just suffer total internal reflection if it is incident at the critical angle $i_c$

Thus, $r_2=i_c$
$\sin i_c=\frac{1}{\mu }=\frac{1}{1.5424}=0.66$ $i_c=si{n}^{-1}\left(0.66\right)={41}^o$

But
$r_1+r_2=A\Rightarrow r_1=A-r_2=A-i_c$
$\Rightarrow r_1=A-r_2=A-i_c={60}^o-{41}^o={19}^o$
From Snell's law,
$\mu =\frac{\sin i}{\sin r_1}$
$\therefore \sin i=\mu \times \sin r_1=1.524\times \sin {19}^o=1.524\times 0.3256=0.4962$
Hence $i={\sin }^{-1}\left(0.4962\right)={30}^o$

Why this question? Note: When light rays travel from optically denser medium to optically rarer medium, at certain angle of incidence, the light ray undergoes total internal reflection. We call this angle of incidence as critical angle.