Question

If the refractive index of water is 4/3 and that of glass is 5/3, then the critical angle of light entering from glass into water will be

• A. ${\sin }^{-1}\left(\frac{4}{5}\right)$
• B. ${\sin }^{-1}\left(\frac{3}{4}\right)$
• C. ${\sin }^{-1}\left(\frac{1}{2}\right)$
• D. ${\sin }^{-1}\left(\frac{2}{3}\right)$

Answer 1

The correct option is A ${\sin }^{-1}\left(\frac{4}{5}\right)$

When a light ray travels from denser to rarer medium, then the angle of incidence for which the refracted ray becomes parallel to the interface$(\angle r={90}^{\circ })$, is known as critical angle for that pair of media.

Here, the denser medium is Glass ${\mu }_g=\frac{5}{3}$

&

the rarer medium is water ${\mu }_w=\frac{4}{3}$

Applying Snell's law at the interface of the two media, we have

${\mu }_g\times Sin{i}_c={\mu }_w\times Sinr$

From definition, we know,

$\angle r={90}^{\circ }$

$\Rightarrow {\mu }_g\times Sin{i}_c={\mu }_w\times Sin{90}^{\circ }$

$\Rightarrow \frac{5}{3}\times Sin{i}_c=\frac{4}{3}$

$\Rightarrow Sin{i}_c=\frac{(\frac{4}{3})}{(\frac{5}{3})}$

$\Rightarrow Sin{i}_c=\frac{4}{5}$

$\Rightarrow i_c=Si{n}^{-1}(\frac{4}{5})$

Hence, the correct answer is OPTION A.