#
The focal length of a Plano convex glass lens is 20cm (mew=1.5).The plane face is silvered .An illuminating object is placed at a distance of 60cm from the lens on its axis along the convex side .Then the. distance (in cm) of the image is

1)20cm 2)30cm 3)40cm 4)12cm

Please find below the solution to the asked query:

First we need tofind the equivalent focal length of a system in which plano convex lens has one surface silvered, we consider the phenomenon taking place in three steps:

- Refraction through first surface (convex surface)
- Reflection through silvered surface (plane surface)
- Refraction through first surface again (convex surface)

$\frac{1}{{f}_{eq}}=\frac{1}{{f}_{convex}}+\frac{1}{{f}_{silvered}}+\frac{1}{{f}_{convex}}\phantom{\rule{0ex}{0ex}}\frac{1}{{f}_{eq}}=\frac{1}{{f}_{convex}}+\frac{1}{\infty}+\frac{1}{{f}_{convex}}\phantom{\rule{0ex}{0ex}}\frac{1}{{f}_{eq}}=\frac{2}{{f}_{convex}}=\frac{2}{20cm}\phantom{\rule{0ex}{0ex}}{f}_{eq}=10cm\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Now,u=-60cm,{f}_{eq}=10cm\phantom{\rule{0ex}{0ex}}Fromlen\text{'}sequation\phantom{\rule{0ex}{0ex}}\frac{1}{{f}_{eq}}=\frac{1}{v}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}\frac{1}{v}=\frac{1}{{f}_{eq}}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\frac{1}{v}=\frac{1}{10}-\frac{1}{60}\phantom{\rule{0ex}{0ex}}v=\frac{10\times 60}{60-10}=12cm\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Hope this information will clear your doubts about the topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards

**
**