Question

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 (in $cm$ )

Answer 1

For lens, using lens maker's formula
$\frac{1}{ f }=\left(\frac{\mu_{2}}{\mu_{1}}-1\right)\left(\frac{1}{ R _{1}}-\frac{1}{ R _{2}}\right)$
$\frac{1}{10}=\left(\frac{1.5}{1}-1\right)\left(\frac{1}{ R }-\frac{1}{- R }\right)$
$\Rightarrow \frac{1}{10}=0.5 \times \frac{2}{ R } $
$\Rightarrow R =10\, cm$

Focal length of equivalent mirror:
$\frac{1}{f_{\text {eq }}}=\frac{1}{10}+\frac{1}{5}+\frac{1}{10}=\frac{2}{5} $
$f_{\text {eq }}=5 / 2 \,cm$
So equivalent mirror is a concave mirror and object must be placed at center of curvature of equivalent mirror, i. e. at distance $5 \,cm$ from its pole, so that its image coincides with it.