Question

(a) What do you understand by the power of a lens? Name one factor on which the power of a lens depends.
(b) What is the unit of power of a lens? Define the unit of power of a lens.
(c) A combination of lenses for a camera contains two converging lenses of focal lengths 20 cm and 40 cm and a diverging lens of focal length 50 cm. Find the power and focal length of the combination.A optical instrument in which the above arrangement of convex lens is used is a convex lens used to burn paper by focusing sunlight.

Answer 1

(a) The power of a lens is a measure of the degree of convergence or divergence of light rays falling on it. It is defined as the reciprocal of the focal length of a lens in metres.
i.e., Power = $\frac{1}{\mathrm{focal}\mathrm{length}\mathrm{of}\mathrm{lens}(\mathrm{in}\mathrm{metre})}$

Therefore, the power of a lens depends on the focal length.

(b) The unit for the power of a lens is a dioptre (D). One dioptre is the power of a lens whose focal length is 1 metre.
(c) Focal length (f₁) of the 1^st converging lens = 20 cm = 0.2 m
Focal length (f₂) of the 2^nd converging lens = 40 cm = 0.4 m
Focal length (f₃) of the diverging lens = -50 cm = -0.5 m (negative sign due to sign convention)
Power = $\frac{1}{\mathrm{focal}\mathrm{length}\mathrm{of}\mathrm{lens}(\mathrm{in}\mathrm{metre})}$

Power (P₁) of 1^st lens = $\frac{1}{0.2}=5\mathrm{D}$

Power (P₂) of 2^nd lens = $\frac{1}{0.4}=2.5\mathrm{D}$

Power (P₃) of diverging lens = $\frac{1}{-0.5}=-2\mathrm{D}$

Net power of the combination = P₁ + P₂+ P₃ = 5 + 2.5 + (-2) = 5.5 D

Net focal length of the combination = $\frac{1}{\mathrm{net}\mathrm{power}}=\frac{1}{5.5}=0.1818\mathrm{m}=18.18\mathrm{cm}$.