Question

An object is placed at a distance of 100cm from a converging lens of focal length 40cm. What is the nature and size of the image formed

Answer 1

bject distance, u=-100 cm
Focal length, f=40 cm
Using the lens formula,
1/v - 1/u=1/f
or 1/v =1/f +1/u
=1/40 -1/100
or v=+ 66.7 cm
m =v/u
= (200/3)/-100
=-0.67

Answer 2

Explanation:

Object distance, u = -100 cm

Focal length, f = +40 cm

We have to find the nature and size of the image formed. Using lens formula :

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

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{40}+\dfrac{1}{(-100)}$

v = 66.7 cm

Magnification of the lens, m = v/u

$m=\dfrac{66.7}{-100}=-0.667$

Since, v is positive. So, the image formed by the converging lens is real and inverted.