Question

Consider the situation described in the previous problem. Where should a point source be placed on the principal axis so that the two images form at the same place?

Answer 1

A point source be placed at the distance of 30 cm on the principal axis, so that the two images form at the same place.

Explanation:

Let the source be set as shown at a distance ' x ' from the lens, so that images produced by both coincide.

To the lens

$\left(\frac{1}{v_{l}}\right)-\left(\frac{1}{-x}\right)=\frac{1}{15}$

$\left(\frac{1}{v_{l}}\right)+\left(\frac{1}{x}\right)=\frac{1}{15}$

$\left(\frac{1}{v_{l}}\right)=\frac{1}{15}+\frac{1}{x}$

$\left(\frac{1}{v_{l}}\right)=\frac{x+15}{15 x}$

$v_{l}=\frac{15 x}{x+15}$

To  the mirror  

u = - (50-x)

f = -10 cm

So,

$\left(\frac{1}{v_{m}}\right)+\frac{1}{-(50-\mathrm{x})}=-\frac{1}{10}$

$\left(\frac{1}{v_{m}}\right)=-\frac{1}{10}+\frac{1}{(50-\mathrm{x})}$

$\left(\frac{1}{v_{m}}\right)=\frac{+10-50+x}{10(50-x)}$

$\left(\frac{1}{v_{m}}\right)=\frac{x-40}{10(50-x)}$

$\left(\frac{1}{v_{m}}\right)=\frac{x-40}{(500-10 x)}$

$500-10 x=v_{m}(x-40)$

$v_{m}=\frac{500-10 x}{(x-40)}$

Since the lens and the reflector are 50 cm apart

$v_{l}-v_{m}=\frac{15 x}{x+15}-\frac{500-10 x}{(x-40)}$

$50=\frac{15 x}{x+15}-\frac{500-10 x}{(x-40)}$

After solving  

We get,

x = 30 cm  

So, place the source 30 cm away from the lens .