Question

A double convex lens made of glass of refractive index 1.5 has its both surfaces of equal radii of curvature of 20 cm each. An object of 5 cm height is placed at a distance of 10 cm from the lens. Find the position, nature and size of the image.

Answer 1

Here n = 1.5 ; R₁ = +20 cm ; R₂ = – 20 cm 

Using lens maker’s formula,

\(\frac{1}{f}\) = \(\frac{1}{20}\)

f = 20 cm

Now , u = -10 cm andf = + 20 cm

h₂ 

Hence a virtual and erect image of height 10 cm is formed at a distance of 20 cm from the lens on the same side as the the object.