A ray of light is reflected by a plane mirror. $\hat{e}_{0}, \hat{e}$ and $\hat{n}$ be the unit vectors along the incident ray, reflected ray and the normal to the reflecting surface respectively. Which of the following gives an expression for ê?

• A

  $\hat{e}_{0}+2\left(\hat{e}_{0} \cdot \hat{n}\right) \hat{n}$
• Correct answer
  B

  $\hat{\boldsymbol{e}}_{0}-2\left(\hat{\boldsymbol{e}}_{0} \cdot \hat{\mathbf{n}}\right) \hat{\mathbf{n}}$
• C

  $\hat{\mathbf{e}}_{0}-\left(\hat{\boldsymbol{\theta}}_{0} \cdot \hat{\mathbf{n}}\right) \hat{\mathbf{n}}$
• D

  \(\mathrm{d}) \hat{\mathrm{e}}_{0}+\left(\hat{e}_{0} \cdot \hat{n}\right) \hat{\mathrm{n}\)

Solution

$\text { (b) According to the question, }$ We know that incident ray, reflected ray and normal lie in the same plane. and angle of incidence $=$ angle of reflection Therefore $\hat{\mathrm{n}}$ will be along the angle bisector of e and $-\hat{e}_{0}$ i.e. $\hat{\mathbf{n}}=\frac{\hat{e}+\left(-\hat{\mathbf{e}}_{0}\right)}{\left|\hat{e}-\hat{\mathbf{e}}_{0}\right|}$ [$\because$ Bisector will along a vector dividing in same ratio as the ratio of sides forming that angle ] But $\hat{\mathbf{n}}$ is a unit vector where $\left|\mathbf{e}-\hat{\mathbf{e}}_{0}\right|=O C$ $\begin{aligned} &=20 P=2|\hat{e}| \\ \cos \theta &=2 \cos \theta \end{aligned}$ Substituting this value in Eq. (i), we get $\hat{\mathbf{n}}=\frac{\hat{\mathbf{e}}-\hat{\mathbf{e}}_{0}}{2 \cos \theta}$ $\begin{array}{l} \hat{e}=\hat{e}_{0}+(2 \cos \theta) \hat{n} \\ \left.\hat{e}=\hat{e}_{0}-2 \hat{n} \cdot \hat{e}_{0}\right) \hat{n}\left[\because \hat{n} \cdot \hat{e}_{0}=-\cos \theta\right] \end{array}$

Hence, the correct answer is option: 'B'.