Question

Dev is observing a small bug using a magnifying lens (convex lens). Initially, he kept the lens at a 10 cm distance from the bug and he saw a magnified upright image of the bug twice its original size. Dev moved the magnifying lens 5 cm closer to the bug. What is the size of the image that he will see?

[0.8 mark]

• A. Twice the original size of the bug
• B. Thrice the original size of the bug
• C. 1/3 of the original size of the bug
• D. 4/3 of the original size of the bug

Answer 1

The correct option is D 4/3 of the original size of the bug

Given:
Initially,
$u=-10cm$
$m=2$

We know that,
Magnification, $m=\frac{Heightofimage\left(I\right)}{Heightofobject\left(O\right)}=\frac{v}{u}$

$2=\frac{V}{-10}$

$v=-20cm$

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

$f=+20cm$

After moving the lens 5 cm closer,
$u\text{'}=-5cm$

Using lens formula,
$\frac{1}{f}=\frac{1}{v\text{'}}-\frac{1}{u\text{'}}$

$v\text{'}=-20/3cm$

$m\text{'}=\frac{v\text{'}}{u\text{'}}=4/3$

$m\text{'}=\frac{Heightofimage(I\text{'})}{Heightofobject\left(O\right)}=\frac{4}{3}$

$I\text{'}=\frac{4}{3}O$