Question

An object is placed at a distance $20cm$ from a thin lens and a real image is formed. The object is now placed at a distance $7.5cm$ from the lens and the image formed is now virtual. If the size of the images in both cases is the same, focal length of the lens is nearly

• A. $11.4cm$
• B. $8.6cm$
• C. $19.2cm$
• D. $13.8cm$

Answer 1

The correct option is D $13.8cm$

Given,

Only convex lens forms both real and virtual image.

Let $f$ be focal length and $m$ be magnification.

For real image, magnification $m=n$

$u=-20$

$v=nu=+20u$

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

$\frac{1}{20n}-\frac{1}{-20}=\frac{1}{f}$

$1+\frac{1}{n}=\frac{20}{f}$ ---- (1)

When virtual image of same size informed magnification is $m=-n$

$u=-7.5$

$v=nu=+7.5u$

$\frac{1}{-7.5n}+\frac{1}{7.5}=\frac{1}{f}$

$1-\frac{1}{n}=\frac{7.5}{f}$---- (2)

Solving (1) and (2) we get,

$f=13.75cm$

Hence, focal length is approx.$13.8cm$