Question

If the incident ray given by vector $\overrightarrow{a}=\hat{i}-\hat{j}$, is incident on a plane mirror having its normal along $\hat{j}$, then the unit vector along the reflected ray is given by :

• A. $\frac{\hat{i}+\hat{j}}{\sqrt{3}}$
• B. $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$
• C. $\hat{i}+\hat{j}$
• D. $\frac{\hat{i}-\hat{j}}{\sqrt{2}}$

Answer 1

The correct option is B $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

Let the unit vector along reflected ray , incident ray and normal be $\widehat{e_r}$ , $\widehat{e_i}$ and $\hat{n}$ respectively.

Given ,
$\overrightarrow{a}=\hat{i}-\hat{j}\Rightarrow \widehat{e_i}=\frac{\hat{i}-\hat{j}}{\sqrt{2}}....\left(1\right)$

Unit vector along reflected light ray is given by

$\widehat{e_r}=\widehat{e_i}-2(\widehat{e_i}\cdot \hat{n})\hat{n}....\left(2\right)$

We find , $(\widehat{e_i}\cdot \hat{n})=\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right).\hat{j}=\frac{-1}{\sqrt{2}}....\left(3\right)$

Using $\left(1\right)$ and $\left(3\right)$ in $\left(2\right)$ we get,

$\widehat{e_r}=\frac{\hat{i}-\hat{j}}{\sqrt{2}}-2\left(\frac{-1}{\sqrt{2}}\right)\hat{j}=\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

Hence, option (b) is the correct answer.

Note:- Emergent ray will be in the same plane as that of the incident ray and the normal.