Question

A thin convex lens of focal length 20 cm is
placed in closed contact with a thin concave
lens of focal length 5 cm. The power of
combination is​

Answer 1

Answer:

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Explanation:

Focal of convex lens(f)=+20cm=20/100=+.20m

44P=1/f=1/.20=100/20=+5D$$

focal length of concave lens(f)=−10cm=−10/100=−.1m

P=1/−.1=−10D

power of combination=5−10=−5D

Answer 2

Given:

A thin convex lens of focal length 20 cm is placed in closed contact with a thin concave lens of focal length 5 cm.

To find:

The power of combination is ?

Calculation:

When 2 lenses are are placed close to one another, the net power of the combination of given as :

$\therefore \: P_{net} = P_{1} + P_{2}$

$\implies \: \dfrac{1}{f_{net}} = \dfrac{1}{f_{1}} + \dfrac{1}{f_{2}}$

$\implies \: \dfrac{1}{f_{net}} = \dfrac{1}{20} + \dfrac{1}{( - 5)}$

$\implies \: \dfrac{1}{f_{net}} = \dfrac{1}{20} - \dfrac{1}{5}$

$\implies \: \dfrac{1}{f_{net}} = \dfrac{1 - 4}{20}$

$\implies \: \dfrac{1}{f_{net}} = \dfrac{ - 3}{20}$

$\boxed{ \bf{ \implies \: f_{net} = \dfrac{ - 20}{3} \: cm }}$

$\implies \: P_{net}=\dfrac{100}{f_{net}(in\:cm)} =\dfrac{100}{( \frac{ - 20}{3})} \:$

$\boxed{ \bf\implies \: P_{net}= - 15 D}$

So, net focal length is -20/3 cm and net power is -15D.