A $2.5$ dioptre lens forms a virtual image which is $4$ times the object placed perpendicularly on the lens. Then the required distance of the object from the lens is .

35

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40

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30

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25

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Solution

The correct option is C $30$
$P = 2.5 d i o p t e r$
$f = \frac{1}{2.5} m = 40 c m$
$\frac{v}{u} = 4$
$l e n s f o r m u l a, \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{f} = \frac{1}{4 u} - \frac{1}{2}$
$\frac{1}{40} = \frac{1 - 4}{4 u} = \frac{- 3}{4 u}$
$u = 30 c m$

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A 2.5 dioptre lens forms a virtual image which is 4 times the object placed perpendicularly on the lens. Then the required distance of the object from the lens is .

The correct option is C 30P=2.5diopterf=12.5m=40cmvu=4lensformula,1v−1u=1f1f=14u−12140=1−44u=−34uu=30cm

A $2.5$ dioptre lens forms a virtual image which is $4$ times the object placed perpendicularly on the lens. Then the required distance of the object from the lens is .

The correct option is C $30$
$P = 2.5 d i o p t e r$
$f = \frac{1}{2.5} m = 40 c m$
$\frac{v}{u} = 4$
$l e n s f o r m u l a, \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{f} = \frac{1}{4 u} - \frac{1}{2}$
$\frac{1}{40} = \frac{1 - 4}{4 u} = \frac{- 3}{4 u}$
$u = 30 c m$