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

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

For the lens,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{15}$

$\Rightarrow v=\frac{15x}{x-15}$ .... (i)

For the mirror,

$u=-(50-x)$

$f=-10\text{cm}$

So, $\frac{1}{V_m}+\frac{1}{(50-x)}=-\frac{1}{10}$

$\Rightarrow \frac{1}{V_m}=\frac{1}{50-x}-\frac{1}{10}$

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

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

So, $V_m=\frac{10(50-x)}{x-40}$ ..... (ii)

Since the lens and mirror are 50 cm apart,

$v-V_m=50$

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

$\Rightarrow (3x^2-120x)-(100x-2x^2-1500+30x)=10(x^2-55x+600)$

$\Rightarrow 5x^2-250x-1500=10x^2-550x+6000$

$\Rightarrow 5x^2-300x+4500=0$

$\Rightarrow x^2-60x+900=0$

$\Rightarrow (x-30{)}^2=0$

$x=30\text{cm}$. So, the source should be placed 30 cm from the lens.