Challenges in optics 2

The question becomes easy if we replace the combination by a single mirror depending on the nature converging or diverging.

In this case $\large{P_{eq}=2P_{lens}+P_{mirror}}$

where P denotes power ,

We know $\large{P_{mirror}=-\frac{1}{f_{mirror}}}$

Also we use lens makers formula.

$\large{\frac{1}{f_{lens}}=(\mu -1)\left( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right)}$

$\large{P_{eq}=2(1.5-1)\left(\frac{1}{-60}-\frac{1}{-20} \right)-\frac{1}{-10}}$

$\large{P_{eq}=-\frac{1}{f_{eq}}=\frac{2}{15}}$

Thus we get $\large{f_{eq}=- \frac{15}{2}}$ thus we get a concave mirror whose focal length is $7.5$ cm.

Thus for a concave mirror , the object should be at the radius of curvature to get the image at the same distance.

Hence the answer is $\large{2f=15~cm=0.15~m}$