The distance between two point sources of light is 24 cm and a converging lens is kept in between two sources. The object distances of two sources from a converging lens of focal length of 9 cm, so that the image distances of two sources coincide

A
6 cm
B
18 cm
C
18 cm or 6 cm
D
24 cm

Solution:

$\begin{aligned} For S_{1} : \frac{1}{v_{1}} - \frac{1}{- x} = \frac{1}{9} \\ ∴ \frac{1}{v_{1}} = \frac{1}{9} - \frac{1}{x} \dots (i) \\ For S_{2} : \frac{1}{v_{2}} - \frac{1}{- (24 - x)} = \frac{1}{9} \dots \\ ∴ \frac{1}{v_{2}} = \frac{1}{9} - \frac{1}{24 - x} \end{aligned} \begin{aligned} v_{1} = - v_{2} \cdot \\ \Rightarrow \frac{1}{v_{1}} = - \frac{1}{v_{2}} \\ ∴ \frac{1}{9} - \frac{1}{x} = \frac{1}{24 - x} - \frac{1}{9} \\ x = 6 c m \end{aligned}$

Solution:

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