Question

A ray parallel to the principal axis is incident at 30° from normal on a concave mirror having a radius of curvature R. The point in principal axis where rays are focussed is Q such that PQ is 

• A. \( \dfrac{R}{2} \)
• B. \( \dfrac{R}{\sqrt{3}} \)
• C. \( \dfrac{2\sqrt{R}}{\sqrt{2}} \)
• D. R(1 - \dfrac1√(3))

Hint:

For mirror problems involving angles: 
• Normal always passes through center of curvature. 
• Use simple trigonometry with reflected rays.

Answer 1

The Correct Option is B

Step 1: For a concave mirror, the normal at point of incidence passes through the center of curvature \( C \).

Step 2: Given angle between incident ray and normal is \( 30^\circ \). Hence, angle between reflected ray and normal is also \( 30^\circ \).

Step 3: Using geometry of reflection and small-angle approximation:
\( \tan 30^\circ = \dfrac{PQ}{PC} \)
\( \dfrac{1}{\sqrt{3}} = \dfrac{PQ}{R} \)
Step 4: Solving:
\( PQ = \dfrac{R}{\sqrt{3}} \)