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=triangle`
B. `x=(f_(1)(f_(2)+d))/(f_(1)+f_(2)-d)`,`y=(triangle)/(f_(1)+f_(2)-d)`
C. `x=(f_(1)f_(2)+d(f_1-d))/(f_(1)+f_(2)-d)`,`y=(triangle(f_(1)-d))/(f_(1)+f_(2)-d)`
D. `x=(f_(1)f_(2)+d(f_(1)-d))/(f_(1)+f_(2)-d)`,`y=0`

Answer

C

Explanation

The image `I_(1)` of parallel rays formed by lens 1 will act as virtual object.
Applying lens formula for lens 2
`implies(1)/(v_(2))=(1)/(f_(1)-d)=(1)/(f_(2))``implies``v_(2)=f_(2)(f_(1)-d)/(f_(2)+f_(1)-d)`
`:"` The horizontal distance of the image I from O is
`x=d+(f_(2)(f_(1)-d))/(f_(2)+f_(1)-d)=(f_(1)f_(2)+d(f_(1)-d))/(f_(1)+f_(2)-d)``m=(v_(2))/(u_(2))=((f_(2)(f_(1)-d))/(f_(1)+f_(2)-d))/(f_(1)-d)=(f_(2))/(f_(1)+f_(2)-d)`
Also `m=h_(2)/(triangle)impliesh_(2)=(trianglexxf_(2))/(f_(1)+f_(2)-d)`
`:'` The y-coordinate `y=triangle-h_(2)`
`=triangle-(trianglef_(2))/(f_(1)_f_(2)-d)=(triangle(f_(1)-d))/(f_(1)+f_(2)-d)`