Question

A ray is incident on a plane surface. If $\hat{i}+\hat{j}-\hat{k}$ represents a vector along the direction of incident ray.$\hat{i}+\hat{j}$ is a vector along normal on incident point in the plane of incident and reflected ray, then vector along the direction of reflected ray is

• A. $-\frac{1}{\sqrt{19}}(-3\hat{i}+3\hat{j}+\hat{k})$
• B. $\frac{1}{\sqrt{19}}(3\hat{i}+3\hat{j}-\hat{k})$
• C. $-\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$
• D. $\hat{k}$

Answer 1

Answer: (C)

According to law of reflection in vector form,
$\widehat{n_2}=\widehat{n_1}-2(\widehat{n_1}.\hat{n})\hat{n}$
Here, $\widehat{n_1}$ = the unit vector along incident ray
$=\frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}$
$\hat{n}$ = unit vector along normal on incident point
$=\frac{\hat{i}+\hat{j}}{\sqrt{2}}$
$\widehat{n_2}$ =unit vector along the direction of reflected ray
Using the formula, we get
$\widehat{n_2}=\frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}-2\left\{\right(\frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}).(\frac{\hat{i}+\hat{j}}{\sqrt{2}}).(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\left)\right\}$
$=\frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}-2\{\frac{1}{\sqrt{6}}+\frac{1}{\sqrt{6}}\}\frac{\hat{i}+\hat{j}}{\sqrt{2}}$
$=\frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}-\frac{4}{\sqrt{6}}\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)$
$=\frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}-\frac{4\hat{i}}{\sqrt{12}}-\frac{4\hat{j}}{\sqrt{12}}$
$=\frac{2\hat{i}+2\hat{j}-2\hat{k}-4\hat{i}-4\hat{j}}{2\sqrt{3}}$ $(\because \sqrt{12}=2\sqrt{3})$
$=\frac{-2\hat{i}-2\hat{j}-2\hat{k}}{2\sqrt{3}}=\frac{-\hat{i}-\hat{j}-\hat{k}}{\sqrt{3}}$
$=-\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$