The object distance u, the image distance v and the magnification m in a lens follow certain linear relations. These are

\frac{1}{u} v e r s u s \frac{1}{v}

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m versus u

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u versus v

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m versus 1/u

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Solution

The correct option is A $\frac{1}{u} v e r s u s \frac{1}{v}$
For a lens $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{v} = \frac{1}{u} + \frac{1}{f} . . . . (i)$
Also $m = \frac{f - v}{f} = 1 - \frac{v}{f} \Rightarrow m = (- \frac{1}{f}) v + 1 . . . (i i)$
On comparing equations (i) and (ii) with y =mx +c.
It is clear that relationship between $\frac{1}{v} v s \frac{1}{u}$ and m vs v is linear.

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The object distance u, the image distance v and the magnification m in a lens follow certain linear relations. These are

The correct option is A 1uversus1vFor a lens 1f=1v−1u⇒1v=1u+1f....(i) Also m=f−vf=1−vf⇒m=(−1f)v+1...(ii) On comparing equations (i) and (ii) with y =mx +c. It is clear that relationship between 1vvs1u and m vs v is linear.