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Question

(C) When an object is placed at the center of curvature of a concave mirror,the image coincides with the object. Thus,the radius of curvature $R = 50

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(C) When an object is placed at the center of curvature of a concave mirror,the image coincides with the object. Thus,the radius of curvature $R = 50 \ cm$.When the mirror is placed in a liquid of refractive index $\mu$,the light rays from the object travel through air and then through the liquid to reach the mirror.The distance of the mirror from the surface of the liquid is $20 \ cm$. The object is at a distance of $40 \ cm$ from the mirror,which means it is at a distance of $40 - 20 = 20 \ cm$ from the surface of the liquid in the air.The light rays from the object travel $20 \ cm$ in air and then enter the liquid. The apparent depth of the object as seen from the liquid is $d' = \frac{d}{\mu} = \frac{20}{\mu}$.The total distance of the object from the mirror as seen by the mirror is $d_{total} = 20 + \frac{20}{\mu}$.Since the image coincides with the object,this total distance must be equal to the radius of curvature $R = 50 \ cm$.$20 + \frac{20}{\mu} = 50$$\frac{20}{\mu} = 30$$\mu = \frac{20}{30} = \frac{2}{3}$ (Wait,let's re-evaluate: The object is at $40 \ cm$ from the mirror. The mirror is at $20 \ cm$ depth. The distance of the object from the liquid surface is $40 - 20 = 20 \ cm$. The apparent distance of this $20 \ cm$ air-gap as seen from inside the liquid is $20 	imes \mu$. So the total distance is $20 + 20\mu = 50 \implies 20\mu = 30 \implies \mu = 1.5 = \frac{3}{2}$.)Therefore,the correct option is $C$.

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