Question

A light ray travelling parallel to the principle axis of a concave mirror strikes the minor at angle of incidence $\theta$. If the radius of curvature of the mirror is $R$, then after reflection, the ray meets the principle axis at distance $d$ from the centre of curvature, then $d$ is

• A. $\frac{R}{2}$
• B. $R(1-\frac{1}{2\cos \theta })$
• C. $\frac{R}{2\cos \theta }$
• D. $\frac{R}{2}(1+\cos \theta )$

Answer 1

Answer: (A)

$\frac{R}{2}$

$\begin{array}{c}\text{So in}△ABC\Rightarrow \\ \frac{d}{\sin \theta }=\frac{R}{\sin (\pi -2\theta )}\left[\text{By}\right(\text{sin rule}\left)\right]\\ \Rightarrow \frac{d}{\sin \theta }=\frac{R}{\sin \left(2\theta \right)}\Rightarrow \frac{d}{\sin \theta }=\frac{R}{2\sin \theta \cos \theta }\\ \cos \sin 2\theta =2\sin \theta \cos \theta )\end{array}$

$d=\frac{R}{2\cos \theta }\Rightarrow$now as $Q\to 0$

$\therefore d=\frac{R}{2}$

Hence obtion 'A is correct.