#
The magnification produced by a convex lens for two different positions of an object are m1 and m2 respectively (m2>m2) .If d is the distance of separation between the two positions of the focal length of the lens is

1)root of m1*m2

2)d/root of m1_m2

3)dm1*m2/m1_m2

4)d/m1_m2

Please find below the solution to the asked query:

I am considering ${m}_{1}>{m}_{2}$. Let us consider for the first

*u*and

*v*are the object and image distance. So we have

$\frac{v}{u}={m}_{1}$

For second case, because of the reversibility, the object and image distance will be

*v*and

*u*respectively. So we have

$\frac{u}{v}={m}_{2}$

Now multiplying above two equation we have

${m}_{1}{m}_{2}=1$

Also we have

$v-u=d\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{1}u-u=d\phantom{\rule{0ex}{0ex}}\Rightarrow u\left({m}_{1}-1\right)=d\phantom{\rule{0ex}{0ex}}\Rightarrow u=\frac{d}{\left({m}_{1}-1\right)}$

Now the focal length of the lens is given by

$f=\frac{uv}{u+v}=\frac{u\times u{m}_{1}}{u+u{m}_{1}}=\frac{u{m}_{1}}{1+{m}_{1}}=\frac{u{\displaystyle \frac{d}{{m}_{1}-1}}}{{m}_{1}+1}=\frac{d{m}_{1}}{{{m}_{1}}^{2}-1}$

Now multiplying ${m}_{2}$ up and down we get

$f=\frac{d{m}_{1}{m}_{2}}{{{m}_{1}}^{2}{{m}_{2}}^{2}-{m}_{2}}=\frac{d}{{m}_{1}{m}_{2}{m}_{1}-{m}_{2}}=\frac{d}{{m}_{1}-{m}_{2}}$

(4) is correct answer.

Hope this information will clear your doubts about this topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards

**
**