For the same statement as above, the ration of the two image sizes for these two positions of the lens is

Options

A. `[(D-d)/(D+d)]^(2)`
B. `[(D+d)/(D-d)]^(2)`
C. `[(D-2d)/(D+2d)]^(2)`
D. `[(D+2d)/(D-2d)]^(2)`

Answer

B

Explanation

If the object is at `u=x_(1)` ,
`m_(1)=(I_(1))/(O)=(D-x_(1))/(x_(1))`
Now, `x_(1)=(1)/(2)(D-d),`
where `d=sqrt(D(D-4f))`
`m_(1)=(D-(D-d)//2)/((D+d)//2)=((D+d)/(D-d))`
Similarly, when the object is at `x_(2)` , the magnification is
`X_(2)=(1)/(2)(D+d)`
`m_(2)=(I_(2))/(O)=(D-x_(2))/(x_(2))=(D-(D+d)//2)/((D+d)//2)=(D=d)/(D+d)`
The ratio of magnification is
`(m_(2))/(m_(1))=((D-d)(D+d))/((D+d)(D-d))=((D-d)/(D+d))^(2)`