Question

A convex lens of focal length 25 cm and a concave lens of focal length 10 cm are placed in close contact with each other. Calculate the lens power of this combination.

Answer 1

Given 

f₁ = 25cm (Focal length of convex lens is positive) 

⇒ f₁ = 0.25m (1m = 100cm)

P (in dioptre) = \(\frac{1}{f(in\,meters)}\)

⇒ Power of convex lens,(F=25 cm) P₁ = \(\frac{1}{f_1}\)

⇒ P₁ = \(\frac{1}{0.25} = \frac{1}{\frac{1}{4}}\)

∴ P₁ = 4 D

f₂ = -10cm (Focal length of concave lens is negative) 

⇒ f₂ = -0.1m (1m = 100cm) 

⇒ Power of the concave lens (F=10 cm), P₂ = \(\frac{1}{f_2}\)

⇒ P₂ = \(-\frac{1}{0.1}=\frac{1}{\frac{1}{10}}\)

∴ P₂ = -10 D. 

Power of combination = P₁ + P₂.

P = 4D – 10D = -6D. 

Lens power of the combination is -6D. Since it is negative therefore the new lens will behave like concave lens having a focal length of 16.6cm.