# 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

Dear Student,

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
$v-u=d\phantom{\rule{0ex}{0ex}}⇒{m}_{1}u-u=d\phantom{\rule{0ex}{0ex}}⇒u\left({m}_{1}-1\right)=d\phantom{\rule{0ex}{0ex}}⇒u=\frac{d}{\left({m}_{1}-1\right)}$
Now the focal length of the lens is given by
$f=\frac{uv}{u+v}=\frac{u×u{m}_{1}}{u+u{m}_{1}}=\frac{u{m}_{1}}{1+{m}_{1}}=\frac{u\frac{d}{{m}_{1}-1}}{{m}_{1}+1}=\frac{d{m}_{1}}{{{m}_{1}}^{2}-1}$
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.

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