Question

A converging lens of focal length 12 cm and a diverging mirror of focal length 7.5 cm are placed 5.0 cm apart with their principal axes coinciding. Where should an object be placed so that its image falls on itself?

Answer 1

Let the object be placed at a distance x cm from the lens (away from the mirror).
For the convex lens (1^st refraction) u = − x, f = − 12 cm

From the lens formula:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v}=\frac{1}{(-12)}+\frac{1}{(-x)}\Rightarrow v=-\left({\displaystyle \frac{12x}{x+12}}\right)$
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 convex mirror,
For the mirror,
$$\begin{gathered} u=-\left(5+\frac{12x}{x+12}\right) \\ =-\left(\frac{17x+60}{x+12}\right) \\ f=-7.5\mathrm{cm} \end{gathered}$$

From mirror equation,
$$\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}=\frac{1}{-7.5}+\frac{x+12}{17x+60} \\ \Rightarrow \frac{1}{v}=\frac{17x+60-7.5}{7.5(17x+60)} \end{gathered}$$
$$\begin{gathered} \Rightarrow v=\frac{7.5(17x+60)}{52.5-127.5x} \\ \Rightarrow v=\frac{250(x+4)}{15x-100} \\ \Rightarrow v=\frac{50(x+4)}{(3x-20)} \end{gathered}$$

Thus, this image is formed towards the left of the mirror.
Again for second refraction in concave lens,
$u=-\left[\frac{5-50(x+4)}{3x-20}\right]$
(assuming that the image of mirror formed between the lens and mirror is 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
Thus, 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.