Question

Two converging lenses, each of focal length 15.0 cm, are placed 40.0 cm apart, and an object is placed 30.0 cm in front of the first lens. Where is the final image formed, and what is the magnification of the system?

Answer 1

Given
Focal length f=15 cm
As the left lens is placed at 40 cm to the left of the right lens. Let place the object at 30 cm in front of this lens we can find the image distance using the formula
1/f=1/p+1/q
q = 1 / (1/f - 1/p) = 1/(1/15 - 1/30) = +30.03cm
q is '+', => the image is real and situated 30.03 cm to the right of the left lens.
This real image becomes the object of the second lens.
So for the right lens p = 40 - 30.03= 9.96 cm, then q is:
q = 1 / (1/15 - 1/9.96 ) = -29.64 cm
q is '-' so the image is virtual
Magnification:
M = (q/p)(q/p)' = (30.03/30)(29.64 /9.96)=2.978

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