A diverging meniscus lens has a focal length of $- 20 c m$ . If the lens is held $10 c m$ from the object, the magnification is

- 0.667

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+ 0.667

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- 2

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+ 2

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Solution

The correct option is B $+ 0.667$

Focal length,F=-20 cm
object distance u=10cm,=-10
Frorn Len's formula
$\begin{matrix} \frac{1}{f} & = \frac{1}{v} - \frac{1}{u} \Rightarrow & - \frac{1}{20} = \frac{1}{v} + \frac{1}{10} \Rightarrow & v = \frac{- 20}{3} c m \end{matrix}$
Magnification, $m = \frac{v}{u}$
$\begin{matrix} = \frac{- 20}{3 x - 10} = \frac{2}{3} = & 0.667 \end{matrix}$
option B is correct

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A diverging meniscus lens has a focal length of −20cm. If the lens is held 10 cm from the object, the magnification is

The correct option is B +0.667Focal length,F=-20 cmobject distance u=10cm,=-10Frorn Len's formula1f=1v−1u⇒−120=1v+110⇒v=−203 cmMagnification,m=vu=−203x−10=23=0.667option B is correct

A diverging meniscus lens has a focal length of $- 20 c m$ . If the lens is held $10 c m$ from the object, the magnification is