Question

Consider the situation shown in the figure. Here $\theta$ is the maximum angle for which the light suffers total internal reflection at the vertical surface. Then $sin\theta$ will be:-

Answer 1

Answer: 00.75

The critical angle for this case is
${\theta }^{{}^{\prime \prime }}={\text{sin}}^{-1}\frac{1}{1\text{.}25}={\text{sin}}^{-1}\frac{4}{5}$
or, $\text{sin}{\theta }^{{}^{\prime \prime }}=\frac{4}{5}$
Since ${\theta }^{{}^{\prime \prime }}=\frac{\pi }{2}-{\theta }^{{}^{\prime }}\text{,}$ we have $\text{sin}{\theta }^{{}^{\prime }}=\text{cos}{\theta }^{{}^{\prime \prime }}=3/5\text{.}$ 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 ${\theta }^{{}^{\prime \prime }}$ 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 ${\sin }^{-1}(3/4)$.