A convex lens of focal length 25 cm and a concave lens of focal length

1.	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» 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.

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»

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.