An object is placed at a distance of 50 , cm | vedclass

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(D) The object is at a distance of $50 \, cm$ from the convex mirror. The plane mirror is placed at a distance of $30 \, cm$ from the object. Since th

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$25$

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(D) The object is at a distance of $50 \, cm$ from the convex mirror. The plane mirror is placed at a distance of $30 \, cm$ from the object. Since the total distance from the object to the convex mirror is $50 \, cm$,the distance from the plane mirror to the convex mirror is $50 - 30 = 20 \, cm$. The plane mirror forms a virtual image of the object at a distance of $30 \, cm$ behind the plane mirror. The distance of this image from the convex mirror is $30 + 20 = 50 \, cm$ behind the mirror. However,the problem states there is no parallax,meaning the images formed by both mirrors coincide. The image formed by the plane mirror is at a distance of $30 \, cm$ behind the mirror. Looking at the geometry,the distance of the image formed by the plane mirror from the pole $P$ of the convex mirror is $v = 30 - 20 = 10 \, cm$ (behind the mirror). Thus,for the convex mirror,$u = -50 \, cm$ and $v = +10 \, cm$. Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$: $\frac{1}{10} + \frac{1}{-50} = \frac{1}{f}$ $\frac{5 - 1}{50} = \frac{1}{f} \Rightarrow \frac{4}{50} = \frac{1}{f} \Rightarrow f = 12.5 \, cm$. The radius of curvature $R = 2f = 2 	imes 12.5 = 25 \, cm$.

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