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A luminous object is placed at a distance of 30 cm from the convex lens of focal length 20 cm . On the other side of the lens, at what distance from the lens a convex mirror of radius of curvature 10 cm be placed in order to have an upright image of the object coincident with it
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50 cm
Using the formula
$\begin{aligned} \frac{1}{v}-\frac{1}{u} & =\frac{1}{f} \\ \frac{1}{v}-\frac{1}{-30} & =\frac{1}{20} \\ v & =60 \mathrm{~cm}\end{aligned}$
If the image is formed at the centre of curvature of the mirror, the coincidence is possible. Then the rays refracting through the lens will fall normally on the convex mirror and retrace their path of form the image at $O$. Hence, the distance between lens and mirror will be $60-10=50 \mathrm{~cm}$.
$\begin{aligned} \frac{1}{v}-\frac{1}{u} & =\frac{1}{f} \\ \frac{1}{v}-\frac{1}{-30} & =\frac{1}{20} \\ v & =60 \mathrm{~cm}\end{aligned}$
If the image is formed at the centre of curvature of the mirror, the coincidence is possible. Then the rays refracting through the lens will fall normally on the convex mirror and retrace their path of form the image at $O$. Hence, the distance between lens and mirror will be $60-10=50 \mathrm{~cm}$.
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