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

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(B) For the image to coincide with the object,the light rays must strike the convex mirror normally. This means the rays must be directed towards the

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(B) For the image to coincide with the object,the light rays must strike the convex mirror normally. This means the rays must be directed towards the center of curvature of the mirror.First,we find the position of the image formed by the convex lens using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Given: $f = +10\,cm$,$u = -15\,cm$.$\frac{1}{v} = \frac{1}{10} + \frac{1}{-15} = \frac{3-2}{30} = \frac{1}{30}$.So,$v = +30\,cm$.This means the lens forms an image at $30\,cm$ from the lens.The convex mirror is placed at the focus of the lens,which is $10\,cm$ from the lens. The distance between the lens and the mirror is $d = 10\,cm$.The distance of the image formed by the lens from the mirror is $MI = v - d = 30\,cm - 10\,cm = 20\,cm$.Since the rays must strike the mirror normally to retrace their path,the point $I$ must be the center of curvature of the convex mirror.Therefore,the radius of curvature $R = MI = 20\,cm$.The focal length of the mirror is $f_m = \frac{R}{2} = \frac{20}{2} = 10\,cm$.

Column $I$	Column $II$
$A$. Concave Mirror	$(p)$ Real image
$B$. Convex Mirror	$(q)$ Virtual image
$C$. Convex Lens	$(r)$ Magnified image
$D$. Concave Lens	$(s)$ Image at infinity

An object is placed at a distance of $15\,cm$ from a convex lens of focal length $10\,cm$. On the other side of the lens,a convex mirror is placed at its focus such that the image formed by the combination coincides with the object itself. The focal length of the convex mirror is.....$cm$

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