Question

Find the distance between two possible positions of a real object in front of a convex lens of focal length 15 cm such that the size of the image is double the size of the object.

• A. 25 cm
• B. 15 cm
• C. 20 cm
• D. 30 cm

Answer 1

The correct option is B 15 cm

Given, focal length of the lens, f = 15 cm
Case I: When the image formed is real and inverted
So, magnification, m = – 2
$\Rightarrow \frac{v}{u}=-2\left(\because m=\frac{v}{u}\right)$$\Rightarrow u=-\frac{v}{2}$Using the lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$\Rightarrow \frac{-1}{2u}-\frac{1}{u}=\frac{1}{15}\Rightarrow u=-22.5cm$

Case II: When the image formed is virtual and erect:
So, magnification, m = +2
$\Rightarrow \frac{v\text{'}}{u\text{'}}=+2$
Using the lens formula,
$\frac{1}{v\text{'}}-\frac{1}{u\text{'}}=\frac{1}{f}$$\Rightarrow \frac{1}{2u\text{'}}-\frac{1}{u\text{'}}=\frac{1}{15}$$\Rightarrow u\text{'}=-7.5cm$
∴ Distance between the two positions of the object = |u'-u| = |-7.5 + 22.5|
=15 cm
Hence, the correct answer is option (2).