Two thin convex lenses of focal lengths `f_(1)` and `f_(2)` are separated by a horizontal distance d (where`dltf_(1)`,`dltf_(2)`) and their centres are displaced by a vertical separation `triangle` as shown in the fig.
Taking the origin of coordinates O, at the centre of the first lens the x and y coordinates of the focal point of this lens system, for a parallel beam of rays coming form the left, are given by: `

Options

A. `x=(f_(1)f_(2))/(f_(1)+f_(2)),y=Delta`
B. `x=(f_(1)(f_(2)+d))/(f_(1)+f_(2)-d),y=(Delta)/(f_(1)+f_(2))`
C. `x=(f_(1)f_(2)+d(f_(1)-d))/(f_(1)+f_(2)-d),y=(Delta(f_(1)-d))/(f_(1)+f_(2)-d)`
D. `x=(f_(1)f_(2)+d(f_(1)-d))/(f_(1)+f_(2)-d),y=(Delta(f_(1)-d))/(f_(1)+f_(2)-d)`

Answer

c.

Explanation

The image `I^(')` of parallel rays formed by lens 1 will act as a virtual object.
Applying lens formula for lens 2,
`(1)/(v)-(1)/(u)=(1)/(f)`
`rArr(1)/(v)-(1)/(f_(1)-d)=(1)/(f_(2))`
`rArr v=(f_(2)(f_(1)-d))/(f_(2)+f_(1)-d)`
The horizontal distance of the image O is
`x=d+(f_(2)(f_(1)-d))/(f_(2)+f_(1)-d)`
`=(df_(2)+df_(1)-d^(2)+f_(2)f_(1)-df_(2))/(f_(2)+f_(1)-d)`
`=(f_(1)f_(2)+d(f_(1)-d))/(f_(2)+f_(1)-d)`
To find teh y-coordinate, we use magnification formula for lens 2,
`m=(v)/(u)=(f_(2)(f_(1)-d)/(f_(1)+f_(2)-d))/(f_(1)-d)=(f_(2))/(f_(1)+f_(2)-d)`. Also
`m=(h^(2))/(Delta)rArr h_(2)=(Deltaxx f_(2))/(f_(1)+f_(2)-d)`
Therefore, the y-coordinate,
`y=Delta-h_(2)`
`=Delta-(Deltaf_(2))/(f_(1)+f_(2)-d)`
`=(Deltaf_(1)+Deltaf_(2)-Deltad-Deltaf_(2))/(f_(1)+f_(2)-d)=(Delta(f_(1)-d))/(f_(1)-f_(2)-d)`