Question

A ray of light is incident on a plane mirror along a vector $(\hat{i}+\hat{j}-\hat{k})$. The normal on the incident point is along $\left(\hat{k}\right)$. The unit vector along the reflected ray will be:

• A. $\frac{1}{\sqrt{2}}(\hat{i}+\hat{j}-\hat{k})$
• B. $\frac{-1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$
• C. $-\hat{i}-\hat{j}$
• D. $\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$

Answer 1

The correct option is D $\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$

Incident ray vector $\left(\overrightarrow{A}\right)$,
$\overrightarrow{A}=\hat{i}+\hat{j}-\hat{k}$
Normal direction vector $\left(\overrightarrow{N}\right)$,
$\overrightarrow{N}=\left(\hat{k}\right)$
After reflection, the component of incident ray parallel to normal gets reversed.

Thus, reflected ray vector $\left(\overrightarrow{R}\right)$,
$\overrightarrow{R}=\left[\right(\hat{i}+\hat{j}\left)\right]-(-\hat{k})=\hat{i}+\hat{j}+\hat{k}$
So, the unit vector along the reflected ray,
$\overrightarrow{R}=\frac{\overrightarrow{R}}{|\overrightarrow{R}|}$
$\Rightarrow \overrightarrow{R}=\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}=\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$
Hence, option $\left(d\right)$ is the correct answer.