Question

Consider the situation shown in figure. Find the maximum angle $\theta$ for which the light suffers total internal reflection at the vertical surface.

• A. $si{n}^{-1}\left(4/5\right).$
• B. $si{n}^{-1}\left(1/4\right).$
• C. $si{n}^{-1}\left(3/4\right).$
• D. None of the above.

Answer 1

The correct option is C

$si{n}^{-1}\left(3/4\right).$

The critical angle for this case is ${\theta }^{\prime \prime }=si{n}^{-1}\frac{1}{1.25}=si{n}^{-1}\frac{4}{5}$
or, $sin{\theta }^{\prime \prime }=\frac{4}{5}.$
Since ${\theta }^{\prime \prime }=\frac{\pi }{2}-{\theta }^{\prime },$ we have sin ${\theta }^{\prime }$ = cos ${\theta }^{\prime \prime }$ = 3/5. From Snell's law,
$\frac{sin\theta }{sin{\theta }^{\prime }}=1.25$
or, sin $\theta$ = 1.25 $\times$ sin ${\theta }^{\prime }$
$=1.25\times \frac{3}{5}=\frac{3}{4}$
$or,\theta =si{n}^{-1}\frac{3}{4}.$
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 $si{n}^{-1}\left(3/4\right).$