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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 focussed is $Q$ such that $\mathrm{PQ}$ is
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Verified Answer
The correct answer is:
$R\left(1-\frac{1}{\sqrt{3}}\right)$
From similar triangles,
$$
\frac{Q C}{\sin 30^{\circ}}=\frac{R}{\sin 120^{\circ}}
$$
or $\quad Q C=R \times \frac{\sin 30^{\circ}}{\sin 120^{\circ}}=\frac{R}{\sqrt{3}}$
Thus $P Q=P C-Q C=R-\frac{R}{\sqrt{3}}=R\left(1-\frac{1}{\sqrt{3}}\right)$
$$
\frac{Q C}{\sin 30^{\circ}}=\frac{R}{\sin 120^{\circ}}
$$
or $\quad Q C=R \times \frac{\sin 30^{\circ}}{\sin 120^{\circ}}=\frac{R}{\sqrt{3}}$
Thus $P Q=P C-Q C=R-\frac{R}{\sqrt{3}}=R\left(1-\frac{1}{\sqrt{3}}\right)$
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