Physics - Geometrical Optics Question with Solution | TestHub

The critical angle for this case is

${\theta }^{″}= {\text{sin}}^{-1}\frac{1}{1\text{.}25}={\text{sin}}^{\mathrm{-1 }}\frac{4}{5}$

or,$\text{sin}{\theta }^{″}=\frac{4}{5}$

Since${\theta }^{″}=\frac{\pi }{2}-{\theta }^{\prime }\text{,}$we have$\text{sin}{\theta }^{\prime }=\text{cos}{\theta }^{″}=3/5\text{.}$From Snell's law,

$\frac{\text{sin }\theta }{\text{sin }{\theta }^{\prime }}=1\text{.}25$

or,$\text{sin}\theta =1\text{.}25\times \text{sin}{\theta }^{\prime }$

$=1\text{.}25\times \frac{3}{5}=\frac{3}{4}$

or,$\theta ={\text{sin}}^{-1}\frac{3}{4}$.

If${\theta }^{″}$is greater than the critical angle,$\theta$will be smaller than this value. Thus, the maximum value of$\theta$, for which total reflection takes place at the vertical surface, is${\sin }^{-1}(3/4)$.