Two lenses of power +10 D and -5 D are placed in contact. Where should a object be held from the lens, so as to obtain a virtual image of magnification 2?

5 cm

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-5 cm

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10 cm

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-10 cm

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Solution

The correct option is D -10 cm
$P_{1} = + 10 D, P_{2} = - 5 D$
$P = P_{1} + P_{2} = 10 - 5 = 5 D$
f $= \frac{100}{5} = 20 c m$
Magnification $= \frac{v}{u} = 2$
$v = 2 u$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (Lens formula)
$\Rightarrow \frac{1}{2 u} - \frac{1}{u} = \frac{1}{f}$
$\Rightarrow \frac{- 1}{2 u} = \frac{1}{f}$
$\Rightarrow u = \frac{- f}{2} = \frac{- 20}{2} = - 10 c m$

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Two lenses of power +10 D and -5 D are placed in contact. Where should a object be held from the lens, so as to obtain a virtual image of magnification 2?

The correct option is D -10 cmP1=+10D,P2=−5DP=P1+P2=10−5=5Df=1005=20cmMagnification=vu=2v=2u1v−1u=1f (Lens formula)⇒12u−1u=1f⇒−12u=1f⇒u=−f2=−202=−10cm