Question

Consider the ray diagram for the refraction given below. The maximum value of angle 6 for which the light suffers total internal reflection at the vertical surface, is

• A. ${\cos }^{-1}\left(\begin{array}{c}\frac{3}{4}\end{array}\right)$
• B. ${\sin }^{-1}\left(\begin{array}{c}\frac{3}{4}\end{array}\right)$
• C. ${\tan }^{-1}\left(\begin{array}{c}\frac{3}{4}\end{array}\right)$
• D. ${\cos }^{-1}\left(\begin{array}{c}\frac{4}{3}\end{array}\right)$

Answer 1

The correct option is B ${\sin }^{-1}\left(\begin{array}{c}\frac{3}{4}\end{array}\right)$

For total internal reflection to occur at the vertical surface, the minimum angle the ray incident on vertical surface can make is $si{n}^{-1}\left(\frac{1}{1.25}\right)$ $=si{n}^{-1}\left(\frac{4}{5}\right)$

Hence, the maximum angle that the refracted ray at the horizontal surface makes with the vertical $={90}^{\circ }-si{n}^{-1}\left(\frac{4}{5}\right)$

Hence, from Snell's law for reflection at horizontal surface,

$sin{\theta }_{max}=1.25\times sin({90}^{\circ }-si{n}^{-1}(\frac{4}{5}\left)\right))$

$=1.25\times cos\left(si{n}^{-1}\right(\frac{4}{5}\left)\right)$

$=\frac{3}{4}$

$⟹{\theta }_{max}=si{n}^{-1}\left(\frac{3}{4}\right)$