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 cm from the lens (away from the mirror).
For the concave lens (I^st refraction) u = − x, f = − 20 cm
From lens formula:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v}=\frac{1}{(-20)}+\frac{1}{(-x)}\Rightarrow v=-\left({\displaystyle \frac{20x}{x+20}}\right)$
Thus, the virtual image due to the first refraction lies on the same side as that of object (A'B').
This image becomes the object for the concave mirror,
For the mirror,
$$\begin{gathered} u=-\left(5+\frac{20x}{x+20}\right) \\ =-\left(\frac{25x+100}{x+20}\right) \\ f=-10\mathrm{cm} \end{gathered}$$

From mirror equation,
$$\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}=\frac{1}{-10}+\frac{x+20}{25x+100} \\ \Rightarrow \frac{1}{v}=\frac{10x+200-25x-100}{250(x+4)} \end{gathered}$$
$$\begin{gathered} \Rightarrow v=\frac{250(x+4)}{100-15x} \\ \Rightarrow v=\frac{250(x+4)}{15x-100} \\ \Rightarrow v=\frac{50(x+4)}{(3x-20)} \end{gathered}$$
Thus, 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"B")
using lens formula,
$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{x}+\frac{1}{{\displaystyle \frac{\left[5-50(x\times 4)\right]}{3x-20}}}=\frac{1}{-20} \end{gathered}$$
⇒ 25x² − 1400x − 6000 = 0
⇒ x² − 56x − 240 = 0
⇒ (x − 60) (x + 4) = 0
So, x = 60 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.