Question

A converging lens and a diverging lens share the same principal axis there focal length f1 and f2 respectively what should be the distance between them so that a beam parallel to principal axis emerge parallel to principal axis

Answer 1

Given :

Focal length of converging lens = f1.

Focal length of diverging lens = f2.

Incident parallel beam emerges parallel to principal axis.

To find :

Distance between both lenses.

Solution :

For first lens :

It is given that incident line is parallel to the converging lens.

For converging lens, focal length of lens is positive.

For diverging lens, focal length of lens is negative.

The lens maker formula for lens is :

$\bf \: \frac{1}{v} - \frac{1}{u } = \frac{1}{f}$

(equation 1)

where

u = Object distance.

v = Image distance

f = Focal length of lens.

Here,

in case of converging lens,

object distance u = ∞.

Putting values in equation 1 :

$\bf \: \frac{1}{v_{c}} - \frac{1}{ \infty } = \frac{1}{f_{c}} \:$

$\bf \: \frac{1}{v_{c}} - 0 = \frac{1}{f_{c}} \\ \bf v_{c} \: = f_{c} \: \:$

so magnitude of image distance in case of converging lens is :

$\bf \: |v_{c}| = f_{c} \:$

in case of diverging lens,

object distance v = ∞.

Putting values in equation 1 :

$\bf \: \frac{1}{ \infty } - \frac{1}{u_{d}} = \frac{1}{f_{d}}$

$\bf \: 0 \: - \frac{1}{u_{d}} = \frac{1}{f_{d}} \:$

$\bf \: u_{d} = - f_{d} \:$

so magnitude of image distance in case of diverging lens is :

$\bf \: |u_{d}| = f_{c} \: \:$

So,

The distance between first and second lens is :

$\bf \: D = | v_{c} | + |u_{d}| \\ \\ \bf \: D = f_{c} + f_{d} \:$

so,

Net distance between both the lenses is equal to the sum of magnitudes of focal lengths of converging and diverging lenses.