A convex lens (of focal length 20, cm) and a | vedclass

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(B) First,we find the image formed by the convex lens using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Given $u = -30\, cm$ and $f =

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$10$

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(B) First,we find the image formed by the convex lens using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Given $u = -30\, cm$ and $f = +20\, cm$,we have $\frac{1}{v} - \frac{1}{-30} = \frac{1}{20} \Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3-2}{60} = \frac{1}{60}$.Thus,$v = +60\, cm$ to the right of the lens.Since the image position does not change when the mirror is removed,the light rays must be striking the mirror normally,meaning the image formed by the lens is at the center of curvature $(C)$ of the concave mirror.The distance between the lens and the mirror is $80\, cm$. The image is at $60\, cm$ from the lens,so the distance of the image from the mirror is $80 - 60 = 20\, cm$.Therefore,the radius of curvature $R = 20\, cm$,which implies the focal length $f_m = \frac{R}{2} = 10\, cm$.$A$ concave mirror produces a virtual image only when the object is placed between the pole $(P)$ and the focus $(F)$.Thus,the maximum distance of the object from the mirror for which a virtual image is formed is equal to the focal length,which is $10\, cm$.

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