Question

An object is placed at a distance of 40 cm from a concave lens of focal length 20 cm. The nature and distance of image from the lens is

• A. Virtual and -14.3 cm
• B. Real and -14.3 cm
• C. Virtual and -13.3 cm
• D. Real and -13.3 cm

Answer 1

The correct option is C Virtual and -13.3 cm

Here, Object distance, u=-40cm
Focal length, f=-20 cm
v=?
Using lens formula
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{-20}+\frac{1}{(-40)}=\frac{-3}{40}v=-40/3\text{Image distance, v=-13.3 cm}\text{Thus, the image is formed at a distance}\text{of 13.3 cm from the concave lens.}\text{Image is virtual and erect.}$