# Two convex lenses of focal length 20 cm are placed co axially with separation of 60 cm between them. Find image of distant object.

(a)40 cm to left of 1st lens

(b)40 cm to right of 1st lens

(c)40 cm to left of second lens

(d)40 cm to right of second lens

The equivalent focal length is
$\large\frac{1}{F} =\frac{1}{f_1}+\frac{1}{f_2} -\frac{d}{f_1f_2}$
$\qquad= \large\frac{1}{20}+\frac{1}{20} -\frac{60}{(20)^2}$
or $F=-20 \;cm$
It is divergent lens. It should be kept at a distance $D= \large\frac{dF}{F}$ behind second lens.
here $D= 60 \times \large\frac{(-20)}{20}$$=-60\;cm$
Thus equivalent divergent lens should be placed at a distance of 60 cm to right of second lens.
The final image is formed at focus of this divergent lens (ie) 20 cm to left of it .
Therefore It is 40 cm to right of second lens.
Hence d is the correct answer.
If F was positive and it was a convergent lens, would the lens be kept at distance D IN FRONT of the second lens, or would it still be placed D behind the second lens, with D=dF/F determining in front or behind the second lens?

Also, why does the F in D=dF/F have to different signs in the calculation?

I guess it's because F is -20 cm and that is done to get negative D to show position behind the second lens. Pretty confusing way to write it I think.