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:

\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\Rightarrow\frac{1}{v}=\frac{1}{( - 20)}+\frac{1}{( - x)}\Rightarrow v=-\left( \frac{20x}{x + 20} \right)\] 
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,
\[u = - \left( 5 + \frac{20x}{x + 20} \right)\]
\[ = - \left( \frac{25x + 100}{x + 20} \right)\]
\[f = - 10 \text{ cm }\]
From mirror equation, 
\[\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\]
\[ \Rightarrow \frac{1}{v} = \frac{1}{- 10} + \frac{x + 20}{25x + 100}\]
\[ \Rightarrow \frac{1}{v} = \frac{10x + 200 - 25x - 100}{250(x + 4)}\] 
\[\Rightarrow v = \frac{250(x + 4)}{100 - 15x}\]
\[ \Rightarrow v = \frac{250(x + 4)}{15x - 100}\]
\[ \Rightarrow v = \frac{50(x + 4)}{(3x - 20)}\]

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,
\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\]
\[ \Rightarrow \frac{1}{x}+\frac{1}{\frac{\left[ 5 - 50 (x \times 4) \right]}{3x - 20}}=\frac{1}{- 20}\]
⇒ 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.