An object is placed at a distance of 15;cm fr | vedclass

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(B) $1$. First,find the image position formed by the convex lens. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with $u = -15\;cm$

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$10$

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(B) $1$. First,find the image position formed by the convex lens. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with $u = -15\;cm$ and $f = 10\;cm$:$\frac{1}{v} - \frac{1}{-15} = \frac{1}{10} \Rightarrow \frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{3-2}{30} = \frac{1}{30}$.Thus,$v = 30\;cm$ from the lens.$2$. The image formed by the lens acts as an object for the mirror. The distance between the lens and the mirror is $30\;cm$. Since the image is formed at $30\;cm$ from the lens,it falls exactly on the mirror.$3$. For the final image to coincide with the object,the rays must strike the mirror normally (i.e.,they must be directed towards the center of curvature of the mirror).$4$. However,the problem states the mirror is placed at its focus. If the rays from the lens are directed towards the focus of the convex mirror,they will reflect back along the same path. Given the distance between the lens and mirror is $30\;cm$,and the image is at $30\;cm$,the rays are incident on the mirror. For the image to coincide with the object,the mirror must reflect the rays back through the lens. This happens if the rays are incident normally on the mirror,meaning the mirror's center of curvature is at the image point. Given the geometry,the focal length of the mirror is $10\;cm$.

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