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
$$\begin{gathered} v-u=d \\ \Rightarrow m_{1}u-u=d \\ \Rightarrow u\left(m_1-1\right)=d \\ \Rightarrow u=\frac{d}{\left(m_1-1\right)} \end{gathered}$$
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. 

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