Question

For the situation shown in the figure below, if incident light ray parallel to principal axis is incident at an angle of $i$ ($R$ is the radius of mirror) then find the distance $\text{QP}$

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

Answer 1

The correct option is D $R(1-\frac{1}{2\cos i})$

Ray diagram for the given problem is drawn below. Point $\text{C}$ is the centre of curvature of the mirror.


From the diagram we can see $\angle QC{A}^{\prime }$ is also $i$.

$\text{QN}$ is the perpendicular bisector on ${\text{A}}^{\prime }\text{C}$.

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

From $\Delta CNQ$,

$\Rightarrow \cos i=\frac{\frac{R}{2}}{CQ}$

$\Rightarrow CQ=\frac{\frac{R}{2}}{\cos i}$

Again from the figure,

$QP=CP-CQ$

$\Rightarrow QP=R-\frac{R}{2\cos i}$

$\Rightarrow QP=R(1-\frac{1}{2\cos i})$

Hence, option (d) is the correct answer.

Why this question: To understand the formation of image by spherical mirror and laws of reflection.