Question

A diverging lens with a magnitude of focal length $25cm$ is placed at a distance of $15cm$from a converging lens of the magnitude of focal length$20cm$. A beam of parallel light falls on the diverging lens. The final image formed is:

• A. Real and at a distance of $40cm$ from the convergent lens.
• B. Virtual and at a distance of $40cm$ from the convergent lens.
• C. Real and at a distance of$6cm$ from the convergent lens.
• D. Real and at a distance of$40cm$ from the divergent lens.

Answer 1

The correct option is A

Real and at a distance of $40cm$ from the convergent lens.

Step 1. Given data,

Focal length of the diverging lens, $f_2=25cm$

Focal length of a converging lens, $f_1=20cm$

Distance between the two lenses, $d=15cm$

Step 2. Finding the distance of the image formed and its nature:

The ray diagram is shown below:

The diverging lens diverges the rays such that the rays coming from inifite appear to meet at focal point $25cm$of the diverging lens.

The distance of object as appeared for the converging lens is,

$\begin{array}{rcl}u& =& -\left(f_2+d\right)\\ & =& -\left(25+15\right)\\ & =& -40cm\end{array}$

Using the lens formula,$\frac{1}{f_1}=\frac{1}{v}-\frac{1}{u}$

Where$f_1$ is the focal length of the converging lens, $u$ is the apparent distance from the converging lens, and $v$ is the distance from the image

$\frac{1}{v}=\frac{1}{f_1}+\frac{1}{u}$

$\frac{1}{v}=\frac{1}{20}-\frac{1}{40}=\frac{1}{40}$

$v=40cm$

The image formed is at a distance of $\mathbf{40}\mathit{c}\mathit{m}$ and is real.

Hence, option A is correct.