Question

An object is placed $30cm$ away from a convex lens of focal length $10cm$ and a sharp image is formed on a screen. Now a concave lens is placed in contact with the convex lens. The screen now has to be moved by $45cm$ to get a sharp image again. The magnitude of focal length of the concave lens is (in $cm$):

• A. 72
• B. 60
• C. 36
• D. 20

Answer 1

The correct option is D 20

Given : $u=-30cm$ $f^{\prime }=10cm$

Convex lens forms an image at point $I^{\prime }$.

Using lens formula $\frac{1}{v^{\prime }}-\frac{1}{u}=\frac{1}{f^{\prime }}$

$\frac{1}{v^{\prime }}-\frac{1}{-30}=\frac{1}{10}$ $⟹$ $v^{\prime }=15$ $cm$

Let the focal length of the combination of the two lenses be $F$ which forms an image at $I$.

$I{I}^{\prime }=45$ cm

$⟹$ $v=v^{\prime }+I{I}^{\prime }=15+45=60$ cm

Using lens formula for the combination $\frac{1}{v}-\frac{1}{u}=\frac{1}{F}$

$\frac{1}{60}-\frac{1}{-30}=\frac{1}{F}$ $⟹$ $F=20$ $cm$

Let the focal length of the concave lens be $f^{\prime \prime }$

Using the focal length formula for combination of lenses $\frac{1}{F}=\frac{1}{f^{\prime }}+\frac{1}{f^{\prime \prime }}$

$\frac{1}{20}=\frac{1}{10}+\frac{1}{f^{\prime \prime }}$ $⟹f^{\prime \prime }=-20$ cm

Hence the magnitude of focal length of the concave lens is $20$ cm