A convex lens of focal length 25 m is placed coaxially in contact with a concave lens of focal length 20 cm .Determine the power of the combination. Will the system be converging or diverging in nature?

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Solution

The net focal length of the combination is
$\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
$= \frac{1}{+ 25} + \frac{1}{- 20}$
$= - \frac{1}{100}$
$⟹ f = - 100 c m$
Power of combination= $P = P_{1} + P_{2} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
$P = \frac{1}{0.25} + \frac{1}{- 0.20} = \frac{100}{25} - \frac{100}{20} = \frac{400 - 500}{100} = - \frac{100}{100} = - 1$
$⟹ P = - 1 D$
Since power is negative, the combination is diverging.

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A convex lens of focal length 25 m is placed coaxially in contact with a concave lens of focal length 20 cm .Determine the power of the combination. Will the system be converging or diverging in nature?

The net focal length of the combination is1f=1f1+1f2=1+25+1−20=−1100⟹f=−100cmPower of combination=P=P1+P2=1f1+1f2P=10.25+1−0.20=10025−10020=400−500100=−100100=−1⟹P=−1DSince power is negative, the combination is diverging.