As shown in the figure, a parallel beam of li | vedclass

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(C) The parallel beam of light is deviated by the prism by an angle $\delta$ given by:$\delta = (\mu - 1) A = (1.5 - 1) 	imes 1.8^{\circ} = 0.5 	ime

Vedclass · Accepted Answer

$3.14 \,mm$

Vedclass · Answer

(C) The parallel beam of light is deviated by the prism by an angle $\delta$ given by:$\delta = (\mu - 1) A = (1.5 - 1) 	imes 1.8^{\circ} = 0.5 	imes 1.8^{\circ} = 0.9^{\circ}$.Converting this angle into radians:$\delta = 0.9^{\circ} 	imes \frac{\pi}{180^{\circ}} 	ext{ rad} = \frac{\pi}{200} 	ext{ rad}$.This beam, after passing through the prism, remains parallel and strikes the concave mirror at an angle of incidence $\delta$ with respect to the principal axis.The rays are focused at a point in the focal plane of the concave mirror.The distance $x$ of this point from the principal axis is given by:$x = f 	imes \delta$, where $f$ is the focal length of the mirror.Given the radius of curvature $R = 40 \,cm$, the focal length $f = \frac{R}{2} = \frac{40}{2} = 20 \,cm$.Substituting the values:$x = 20 \,cm 	imes (0.9^{\circ} 	imes \frac{\pi}{180^{\circ}}) = 20 	imes \frac{\pi}{200} \,cm = \frac{\pi}{10} \,cm = 0.314 \,cm = 3.14 \,mm$.

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