(a) What is meant by ‘power of a lens’?(b) State and define the S.I. unit of power of a lens.(c) 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.

(a) The 'power of a lens' (P) is defined as, the ability of a lens to converge or diverge a beam of light falling on it.

Also, the power of a lens is the reciprocal of its focal length.

That is, $P=\frac{1}{f}$

(b) The S.I unit of power of lens is dioptre, which is denoted by $D$.

$1D=\frac{1}{1m}=1{m}^{-1}$

Here, 1m is the focal length of the lens

Thus, 1 Dioptre is the power of the lens having a focal length of 1m.

(c) Given,

Focal length of convex lens = ${f}_{1}=+25cm$ $=+(\frac{25}{100})m$ (converted cm to m)

Focal length of concave lens = ${f}_{2}=-10cm$ $=-(\frac{10}{100})m$ (converted cm to m)

To find = Power of the combination of lenses.

Solution

We know that power of a lens is the reciprocal of its focal length.

Therefore, the power of a convex lens of focal length 25 cm will be-

${P}_{1}=\frac{1}{{f}_{1}}$

${P}_{1}=\frac{1}{\frac{25}{100}}$

${P}_{1}=\frac{100}{25}$

${P}_{1}=+4D$

Also, the power of a concave lens of focal length 10 cm will be-

${P}_{2}=\frac{1}{{f}_{2}}$

${P}_{2}=\frac{1}{(-\frac{10}{100})}$

${P}_{2}=(-\frac{100}{10})$

${P}_{2}=-10D$

Now, power of the combination of lens is given as-

$P={P}_{1}+{P}_{2}$

$P=4D-10D$

$P=-6D$

Hence, the lens power of this combination will be -6D.

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Updated on: 10-Oct-2022

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