Question

The refractive index of water is $\frac{4}{3}$ and that of glass is $\frac{5}{3}$. What will be the critical angle for the ray of light entering the water from the glass?

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

Answer 1

The correct option is A

${\sin }^{-1}\left(\frac{4}{5}\right)$

Step 1: Given data

The refractive index of water is $\frac{4}{3}$

The refractive index of glass is $\frac{5}{3}$

Step 2: Formula used

Snell's law: This law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for any two given media.

Mathematically

$\frac{\sin {\mathrm{i}}_{\mathrm{c}}}{\sin {\mathrm{r}}_{\mathrm{c}}}{=}^{\mathrm{a}}{\mathrm{\mu }}_{\mathrm{b}}$

Here,

$\sin {\mathrm{i}}_{\mathrm{c}}$ is the angle of incidence.

$\sin {\mathrm{r}}_{\mathrm{c}}$ is the angle of refraction.

${}^{\mathrm{a}}{\mathrm{\mu }}_{\mathrm{b}}$ is the relative refractive index of the medium b with respect to medium a.

Step 3: Calculation

By substituting the values,

$\sin i_c=\frac{{\mu }_w}{{\mu }_g}$

Here,

${\mathrm{\mu }}_{\mathrm{w}}$ is the refractive index of water.

${\mathrm{\mu }}_{\mathrm{g}}$ is the refractive index of glass.

$$\begin{gathered} \sin {\mathrm{i}}_{\mathrm{c}}=\frac{{\displaystyle \frac{4}{3}}}{{\displaystyle \frac{5}{3}}} \\ \sin {\mathrm{i}}_{\mathrm{c}}=\frac{4}{5} \\ {\mathrm{i}}_{\mathrm{c}}={\sin }^{-1}\left(\frac{4}{5}\right) \end{gathered}$$

Therefore, the correct option is (A) ${\sin }^{-1}\left(\frac{4}{5}\right)$.