Answer to Question #285048 in Optics for Tom

Question

A convex lens of focal length 6cm is at distance of 10cm from a screen placed at right angles to axis of lens.A diversing lens of 12cm focal length is palced co-axial between convex lens and screen so that image of object at 24cm from convex lens is formed on secm.The distance between them?

Answer 1

Solution:

The equation for thin lens is

$\begin{aligned} &\frac{1}{f}=\frac{1}{p}+\frac{1}{q} \\ &\Rightarrow \frac{1}{q}=\frac{1}{f}-\frac{1}{p} \\ &\Rightarrow \frac{1}{q}=\frac{1}{6 \mathrm{~cm}}-\frac{1}{24 \mathrm{~cm}} \\ &\Rightarrow q=8.0 \mathrm{~cm} \end{aligned}$

This image is object of diverging lens

let us suppose the distance between the two lenses is d then object distance of diverging lens is

$p=8-d$

Image distance is

$q=10-d$

Use lens equation for diverging lens

$\begin{aligned} &\frac{1}{f}=\frac{1}{(d-8)}+\frac{1}{(10-d)} \\ &\Rightarrow \frac{1}{-12 \mathrm{~cm}}=\frac{2}{-d^{2}+18 d-80} \\ &\Rightarrow d^{2}-18 d+80=12 \mathrm(18-2 d) \\ &\Rightarrow d^{2}-18 d+80=216-24 d \\ &\Rightarrow d^{2}+6 d-136=0 \end{aligned}$

On solving quadratic equaiton, we get,

$d=9.04 \mathrm{~cm}$