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Lens Maker's Formula

The magnifica...

Question

The magnifications produced by a convex lens for two different positions of an object are $m_{1}$ and $m_{2}$ respectively $(m_{1} > m_{2})$ . If d is the distance of separation between the two positions of the object then the focal length of the lens is

A

\sqrt{m_{1} m_{2}}

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B

\frac{d}{\sqrt{m_{1} - m_{2}}}

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C

\frac{d m_{1} m_{2}}{m_{1} - m_{2}}

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D

\frac{d}{m_{1} - m_{2}}

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Solution

The correct option is D $\frac{d}{m_{1} - m_{2}}$
Given: The magnifications produced by a convex lens for two different positions of an object are $m_{1}$ and $m_{2}$ respectively ( $m_{1} > m_{2}$ ). If $d$ is the distance of separation between the two positions of the object
To find the focal length of the lens
Solution:
Separation between object and image for first position,
$D = v + u$
where $u, v$ are the object distance and image distance, respectively.
So, magnification
$m_{1} = \frac{v}{u} ⟹ v = u m_{1} . . . . (i)$
When the lens is at second position,
d=v-u
So, $m_{2} = \frac{u}{v} ⟹ u = v m_{2} . . . . . . (i i)$
So, $m_{1} m_{2} = 1$
From lens equation and using eqn (i), we get
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} ⟹ \frac{1}{f} = \frac{1}{u m_{1}} + \frac{1}{u} ⟹ \frac{1}{f} = \frac{1 + m_{1}}{u m_{1}} ⟹ u = \frac{f (m_{1} + 1)}{m_{1}} . . . . . . . . . (i i i)$
From lens equation and using eqn (ii), we get
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} ⟹ \frac{1}{f} = \frac{1}{v m_{2}} + \frac{1}{v} ⟹ \frac{1}{f} = \frac{1 + m_{2}}{v m_{2}} ⟹ v = \frac{f (m_{2} + 1)}{m_{2}} . . . . . . . . . (i v)$
substituting the values of eqn(iii) and(iv) in the following equation, we get
$d = v - u ⟹ d = \frac{f (m_{2} + 1)}{m_{2}} - \frac{f (m_{1} + 1)}{m_{1}} ⟹ d = f (\frac{m_{1} (m_{2} + 1) - m_{2} (m_{1} + 1)}{m_{1} m_{2}}) ⟹ f = \frac{d m_{1} m_{2}}{m_{1} m_{2} + m_{1} - m_{1} m_{2} - m_{2}} ⟹ f = \frac{d m_{1} m_{2}}{m_{1} - m_{2}}$
but $m_{1} m_{2} = 1$
Therefore the focal length becomes,
$f = \frac{d}{m_{1} - m_{2}}$

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