If the image formed by a thin convex lens of power $P$ has magnification $m$ then image distance $v$ is

v = \frac{1 - m}{P}

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v = \frac{1 + m}{P}

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v = \frac{m}{P}

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v = \frac{1 + 2 m}{P}

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Solution

The correct option is A $v = \frac{1 - m}{P}$
Using lens formula :
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
where focal length of the lens $f = \frac{1}{P}$
So we get, $\frac{1}{v} - \frac{1}{u} = P$
Or $1 - \frac{v}{u} = P v$
Also, magnification of the lens $m = \frac{v}{u}$
$∴$ $1 - m = P v$
$⟹ v = \frac{1 - m}{P}$

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If the image formed by a thin convex lens of power P has magnification m then image distance v is

The correct option is A v=1−mPUsing lens formula :1v−1u=1fwhere focal length of the lens f=1PSo we get, 1v−1u=POr 1−vu=PvAlso, magnification of the lens m=vu∴ 1−m=Pv⟹ v=1−mP

If the image formed by a thin convex lens of power $P$ has magnification $m$ then image distance $v$ is