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As shown in the figure, a parallel beam of light incidents on the upper part of a prism of angle $1.8^{\circ}$ and material of refractive index 1.5. The light emerging out from the prism falls on a concave mirror of radius of curvature $40 \mathrm{~cm}$. This distance of the point from the principal axis of the mirror where the light rays are focussed after reflection from the mirror is
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Verified Answer
The correct answer is:
1.57 mm
Parallel beam is deviated by prism by angle,
$$
\begin{aligned}
\delta & =(\mu-1) A=(1.5-1) \times 1.8^{\circ}=0.9^{\circ} \\
& =0.9^{\circ} \times \frac{\pi}{180^{\circ}} \text { radian }
\end{aligned}
$$
This beam from prism is also parallel and is converged at a point in focal plane by concave mirror.
Distance,
$$
\begin{aligned}
x & =F \times \delta=\frac{R}{2} \times \delta=\frac{20}{2} \times 0.9^{\circ} \times \frac{\pi}{180^{\circ}} \\
& =0.157 \mathrm{~cm}=1.57 \mathrm{~mm}
\end{aligned}
$$
$$
\begin{aligned}
\delta & =(\mu-1) A=(1.5-1) \times 1.8^{\circ}=0.9^{\circ} \\
& =0.9^{\circ} \times \frac{\pi}{180^{\circ}} \text { radian }
\end{aligned}
$$
This beam from prism is also parallel and is converged at a point in focal plane by concave mirror.
Distance,
$$
\begin{aligned}
x & =F \times \delta=\frac{R}{2} \times \delta=\frac{20}{2} \times 0.9^{\circ} \times \frac{\pi}{180^{\circ}} \\
& =0.157 \mathrm{~cm}=1.57 \mathrm{~mm}
\end{aligned}
$$
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