Optics Question 792
A ray parallel to principal axis is incident at $ 30{}^\circ $ from normal on concave mirror having radius of curvature R. The point on principal axis where rays are focused is F such that PF is
Options:
A) $ \frac{R}{2} $
B) $ \frac{R}{\sqrt{3}} $
C) $ \frac{2\sqrt{R}-\sqrt{R}}{\sqrt{2}} $
D) $ R( 1-\frac{1}{\sqrt{3}} ) $
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Answer:
Correct Answer: D
Solution:
[d] From similar triangles, $ \frac{QC}{\sin 30{}^\circ }=\frac{R}{\sin 120{}^\circ } $
or $ QC=R\times \frac{\sin 30{}^\circ }{\sin 120{}^\circ }=\frac{R}{\sqrt{3}} $
Thus $ PQ=PC-QC=R-\frac{R}{\sqrt{3}}=R( 1-\frac{1}{\sqrt{3}} ) $
BETA
