Question

Two thin convex lenses of focal lengths f1 and f2 are separated by a horizontal distance d (where d < f1, d < f2) and their centers are displaced by a vertical separation Δ as shown in the figure.

Taking the origin of coordinates, O at the center of the first lens, the x and y-coordinates of the focal point of this lens system, for a parallel beam of rays coming from the left, are given by

• (A) $x=\frac{f_1{f}_2}{f_1+f_2},Y=\Delta$
• (B) $x=\frac{f_1\left(f_2+d\right)}{f_1+f_2-d},Y=\frac{\Delta }{f_1+f_2}$
• (C) $x=\frac{f_1{f}_2+d\left(f_1-d\right)}{f_1+f_2-d},Y=\frac{\Delta \left(f_1-d\right)}{f_1+f_2-d}$
• (D) $x=\frac{f_1{f}_2+d\left(f_1-d\right)}{f_1+f_2-d},Y=0$

Answer 1

The Correct Option is C

Explanation:
Given:Focal lengths f1 > d and f2 > d Since f1 > d and f2 > d, the focus of both the lenses L1 and L2 lies on the right-hand side of lens L2.The complete ray diagram is shown below:

The image formed by L1 will act as a virtual object for the second lens L2.For second lens:Object distance u = f1 - dfocal length = f2Using the lens formula , image distance v is$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$\Rightarrow \frac{1}{v}=\frac{1}{f_2}+\frac{1}{\left(f_1-d\right)}$$\Rightarrow v=\frac{f_2\left(f_1-d\right)}{f_1+f_2-d}$The x-coordinate of focal point A of the lens system is$x=d+v$$\Rightarrow x=d+\frac{f_2\left(f_1-d\right)}{f_1+f_2-d}$$\Rightarrow x=\frac{f_1{f}_2+d(f1-d)}{f_1+f_2-d}$The linear magnification produced by the second lens L is$m=\frac{v}{u}$$\Rightarrow m=\left(\frac{\frac{f_2\left(f_1-d\right)}{f_1+f_2-d}}{\left(f_1-d\right)}\right)$$\Rightarrow m=\frac{f_2}{f_1+f_2-d}$The y- coordinate of the focal point is$y=\Delta -m\Delta$$\Rightarrow y=\Delta -\frac{f_2\Delta }{f_1+f_2-d}$$\Rightarrow y=\frac{\Delta \left(f_1-d\right)}{\left(f_1+f_2-d\right)}$Therefore, the x and y -coordinates of the focal point of the lens system are$x=\frac{f_1{f}_2+d\left(f_1-d\right)}{f_1+f2-d},y=\frac{\Delta \left(f_1-d\right)}{f_1+f_2-d}$ respectivelyHence, the correct option is (C).