Question

An object of size 3.0 cm is placed 14cm in front of a concave lens of focal length 21cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

Answer 1

Size of the object,h₁=3cm
Object distance,u=-14cm
Focal length of the concave lens,f=-21cm
Image distance=v
According to the lens formula, we have the relation:

\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

=\(\frac{1}{v}=-\frac{1}{21}-\frac{1}{14}\)

=-2\(-\frac{3}{42}\)=\(-\frac{5}{42}\) ∴v=\(-\frac{42}{5}\)=-8.4cm
Hence,the image is formed on the other side of the lens,8.4cm away from it. The negative sign shows that the image is erect and virtual.The magnification of the image is given as:m=Image height(h₂)/Object height(h1)=\(\frac{v}{u}\) ∴h₂=\(\frac{-8.4}{-14}\)×3=0.6×3=1.8cm
Hence, the height of the image is 1.8cm.If the object is moved further away from the lens, then the virtual image will move toward the focus focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.