The lateral magnification of the lens with an object located at two different position `u_(1)` and `u_(2)` are `m_(1)` and `m_(2)` , respectively. Then the focal length of the lens is

Options

A. `f=sqrt(m_(1)m_(2))(mu_(2)-mu_(1))`
B. `f=sqrt(m_(1)m_(2))(u_(2)-u_(1))`
C. `((u_(2)-u_(1)))/(sqrt(m_(1)m_(2)))`
D. `((u_(2)-u_(1)))/((m_(2))^(-1)-(m_(1))^(-1))`

Answer

d.

Explanation

For lens `(1)/(f)=(1)/(v_(1))-(1)/(u_(1))`
`(u_(1))/(f)=(u_(1))/(v_(1))-1rArrm_(1)=(v_(1))/(u_(1))=(f)/(u_(1)+f)`
And `m_(2)=(f)/(u_(2)+f)`
`(1)/(m_(2))-(1)/(m_(1))=((u_(2)-mu_(1)))/(f) rArr f=((mu_(2)-mu_(1)))/((m_(2))^(-1)-(m_(1))^(-1))`