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. `(d(f_(1)-d)+f_(1)f_(2))/((f_(1)+f_(2)-d))`
B. `(f_(1)f_(2))/((f_(1)+f_(2)-d))`
C. `(d(f_(1)-d))/((f_(1)+f_(2)-d))`
D. `(2d(f_(1)+d)-f_(1)f_(2))/((f_(1)+f_(2)-d))`

Answer

a

Explanation

The parallel rays will be focussed at the focal point of the first lens. The first image lies at I, at a distance `f_(1)` from the origin. This image`I_(1)` will act as an object for refraction through the second lens. The object distance for the second, `u=(f_(1)-d)`.
From lens equation, `(1)/(upsilon)-(1)/(+(f_(1)-d))=(1)/(f_(2))`
`upsilon=(f_(2)(f_(1)-d))/((f_(1)+f_(2)-d))`
Hence, the x-coordinate of final image `I_(2)` is
`x=d+u=(d+f_(2)(f_(1)-d))/((f_(1)+f_(2)-d))=(d(f_(1)-d)+f_(1)f_(2))/((f_(1)+f_(2)-d))`
Imagine an arrow tip is at `f_(1):` its image from lens `f_(1)` is the final image.
Magnification, `m+(upsilon)/(u)=(I_(2))/(O)=(I_(2))/(Delta)`
`I_(2)=((upsilon)/(u))Delta=(f_(2))/((f_(1)+f_(2)-d))Delta`
Thus, y-coordinate of tip of `I_(2)` is
`Delta-I_(2)=Delta[1-(f_(1))/((f_(1)+f_(2)-d))]=((f_(1)-d)Delta)/((f_(1)+f_(2)-d))`