Question

A beam of light propagates through a medium $1$ and falls onto another medium $2$, at an angle ${\alpha }_1$ as shown in Fig. After that, it propagates in medium $2$ at an angle ${\alpha }_2$ as shown. The light's wavelength in medium $1$ is ${\lambda }_1$. What is the wavelength of light in medium $2$?

• A. $\frac{\sin {\alpha }_1}{\sin {\alpha }_2}{\lambda }_1$
• B. $\frac{\sin {\alpha }_2}{\sin {\alpha }_1}{\lambda }_1$
• C. $\frac{\cos {\alpha }_1}{\cos {\alpha }_2}{\lambda }_1$
• D. $\frac{\cos {\alpha }_2}{\cos {\alpha }_1}{\lambda }_1$

Answer 1

Answer: (B)

$\frac{\sin {\alpha }_2}{\sin {\alpha }_1}{\lambda }_1$

(B)

Let $n_1$ be the refractive index of medium $1$ and $n_2$ is that of medium $2$.

Therefore bu snell's law,

$n_1\sin {\alpha }_1=n_2\sin {\alpha }_2$

$\frac{\sin {\alpha }_1}{\sin {\alpha }_2}=\frac{n_2}{n_1}$

$n=\frac{C}{V}$ , $C\to$ Velocity of light in vacuum

$V\to$ Velocity of light in medium

$\frac{\sin {\alpha }_1}{\sin {\alpha }_2}=\frac{\frac{C}{V_2}}{\frac{C}{V_1}}=\frac{V_1}{V_2}$

$V=v\lambda$ $v\to$ Frequency

There is no change in frequency when light travels from one medium to another.

$\therefore$ $V\alpha \lambda$

$\frac{\sin {\alpha }_1}{\sin {\alpha }_2}=\frac{{\lambda }_1}{{\lambda }_2}$

$\therefore$ ${\lambda }_2=\frac{\sin {\alpha }_2}{\sin {\alpha }_1}.{\lambda }_1$