Physics - Geometrical Optics Question with Solution | TestHub

$\text{ (b) According to the question, }$ 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. $\hat{\mathbf{n}}=\frac{\hat{e}+\left(-{\hat{\mathbf{e}}}_0\right)}{|\hat{e}-{\hat{\mathbf{e}}}_0|}$ [$\because$ 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 $\left|\mathbf{e}-{\hat{\mathbf{e}}}_0\right|=OC$ $\begin{array}{rl}& =20P=2|\hat{e}|\\ \cos \theta & =2\cos \theta \end{array}$ Substituting this value in Eq. (i), we get $\hat{\mathbf{n}}=\frac{\hat{\mathbf{e}}-{\hat{\mathbf{e}}}_0}{2\cos \theta }$ $\begin{array}{l}\hat{e}={\hat{e}}_0+(2\cos \theta )\hat{n}\\ \hat{e}={\hat{e}}_0-2\hat{n}\cdot {\hat{e}}_0)\hat{n}\left[\because \hat{n}\cdot {\hat{e}}_0=-\cos \theta \right]\end{array}$