Question

Two thin lenses of power P1= +0.5 D and P2= -3 D respectively are
placed in contact. The focal length of the combination will be A) 0.4m
B) - 40 cm
C) +40 cm
D) 2.5 m

Answer 1

Given,

$\to f_1 = \dfrac{100}{0.5} = 200\: cm\:\:\:\: f_2 = \dfrac{100}{-3} \: cm$

We know that,

$\implies\rm P_{net} = P_1 + P_2$

$\implies \bf \dfrac{1}{f_{net}} = \dfrac{1}{f_1} + \dfrac{1}{f_2}$

$\implies \rm \dfrac{1}{f_{net}} =\dfrac{1}{200} + \dfrac{-3}{100}$

$\implies \rm\dfrac{1}{f_{net}} =\dfrac{1}{200} - \dfrac{3}{100}$

$\implies \rm \dfrac{1}{f_{net}}= \dfrac{100-600}{200\times 100}$

$\implies \rm \dfrac{1}{f_{net}}= \dfrac{-500}{200\times 100} =\dfrac{-5}{200}$

$\implies \bf f_{net} = -40\: cm$

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Answer 2

Explanation:

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Given,

$\to f_1 = \dfrac{100}{0.5} = 200\: cm\:\:\:\: f_2 = \dfrac{100}{-3} \: cm$

We know that,

$\implies\rm P_{net} = P_1 + P_2$

$\implies \bf \dfrac{1}{f_{net}} = \dfrac{1}{f_1} + \dfrac{1}{f_2}$

$\implies \rm \dfrac{1}{f_{net}} =\dfrac{1}{200} + \dfrac{-3}{100}$

$\implies \rm\dfrac{1}{f_{net}} =\dfrac{1}{200} - \dfrac{3}{100}$

$\implies \rm \dfrac{1}{f_{net}}= \dfrac{100-600}{200\times 100}$

$\implies \rm \dfrac{1}{f_{net}}= \dfrac{-500}{200\times 100} =\dfrac{-5}{200}$

$\implies \bf f_{net} = -40\: cm$

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