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# A convex lens of focal length $25\,cm$ and a concave lens of focal length $10\,cm$ are placed in contact with each other.(a) What is the power of this combination?(b) What is the focal length of this combination?(c) Is this combination converging or diverging?

Last updated date: 12th Aug 2024
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Hint:In order to solve this question, we will use the general formula of finding the net focal length of combination of lenses and net power of combination of lenses. The focal length of a convex lens is positive since it’s a converging lens and the focal length of a concave lens is negative because concave lens is a diverging lens.

Formula used:
Power and focal length of a lens is related as $P = \dfrac{1}{f}$ .

Complete step by step answer:
According to the question, Focal length of convex lens is ${f_{convex}} = + 25cm = + 0.25m$
Power of this lens will be ${P_{convex}} = \dfrac{1}{{0.25}}$
${P_{convex}} = 4D$
And, focal length of concave lens is ${f_{concave}} = - 10cm = - 0.1m$
Power of this lens will be ${P_{concave}} = - \dfrac{1}{{0.1}}$
${P_{concave}} = - 10D$

(a) Net power of the combination can be calculated as
${P_{net}} = {P_{convex}} + {P_{concave}}$
Putting the values of parameters we get,
${P_{net}} = - 10 + 4$
$\therefore {P_{net}} = - 6\,D$

(b) Net focal length of the combination can be found as
Since net power we have calculated is ${P_{net}} = - 6D$
Then, net focal length can be written as ${f_{net}} = \dfrac{1}{{{P_{net}}}}$
$\Rightarrow {f_{net}} = - \dfrac{1}{6}$
$\therefore {f_{net}} = - 0.1666\,m = - 16.66\,cm$
Hence, the focal length of a combination of such systems is $- 16.66\,cm$.

(c) Since, the net focal length of the combination of the system is $- 16.66\,cm$ which is negative and the focal length of concave lens is negative which is a diverging lens hence, the system will behave as a diverging lens.

Note: It should be remembered that, the unit of power is Dioptre and denoted by $D$.$1D$ is the ratio of lens having focal length of $1m$ and basic unit of conversions used are $1m = 100cm$ and if two lenses were kept at a distance of x meter then net focal length will be calculated by using the formula $\dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} - \dfrac{x}{{{f_1}{f_2}}}.$