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 m1>m2. Let us consider for the first u and v are the object and image distance. So we have
vu=m1
For second case, because of the reversibility, the object and image distance will be v and u respectively.  So we have
uv=m2
Now multiplying above two equation we have
m1m2=1
Also we have
v-u=dm1u-u=dum1-1=du=dm1-1
Now the focal length of the lens is given by 
f=uvu+v=u×um1u+um1=um11+m1=udm1-1m1+1=dm1m12-1
Now multiplying m2 up and down we get
f=dm1m2m12m22-m2=dm1m2m1-m2=dm1-m2
(4) is correct answer. 

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