Question

A diverging lens of focal length 20 cm and a converging mirror of focal length 10 cm are placed coaxially at a separation of 5 cm. Where should an object be placed so that a real image is formed at the object itself ?

Answer 1

Let the object be placed at a distance x from the lens farther away from the mirror.

For the concave lens (1st refraction)

u = -x, f = -20 cm

From lens formula

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

$\Rightarrow \frac{1}{v}=\frac{1}{(-20)}+\frac{1}{(-x)}$

$\Rightarrow v=-\left(\frac{20x}{x+20}\right)$

So, the virtual image due to the first refraction lies same side as that of object. $\left(A^{{}^{\prime }}{B}^{{}^{\prime }}\right)$

This image becomes the object for the concave mirror,

For the mirror,

$u=-(5+\frac{20x}{x+20})$

$=-\left(\frac{25x+100}{x+20}\right)$

$f=-10\text{cm}$

From mirror equation,

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

$\Rightarrow \frac{1}{v}=\frac{1}{-10}+\frac{x+20}{25x+100}$

$=\frac{10x+200-25x-100}{250(x+4)}$

$\Rightarrow v=\frac{250(x+4)}{100-15x}$

$=-\frac{250(x+4)}{15x-100}$

$=-\frac{50(x+4)}{(3x-20)}$

So, this image is formed towards left of the mirror.

Again for second refraction in concave lens,

$u=-\left[\frac{5-50(x+4)}{3x-20}\right]$ (assuming that image of mirror is formed between the lens and mirror $3x-20$)

v = +x (since, the final image is produced on the object $A^{\prime \prime }{B}^{\prime \prime }$) using lens formula,

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

$\Rightarrow \frac{1}{x}+\frac{1}{5-50(x\times 4)}=\frac{1}{-20}$

$3x-20$

$\Rightarrow 25x^2-1400x-6000=0$

$\Rightarrow x^2-56x-240=0$

$\Rightarrow (x-60)(x+4)=0$

So, $x=60\text{m}$

The object should be placed at a distance 60 cm from the lens farther away from the mirror so that the final image is formed on itself.