Two-lens systems. In the figure, stick figure O (the object) stands on the commo

Question

Two-lens systems. In the figure, stick figure O (the object) stands on the common central axis of two thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closer to O, which is at object distance p1. Lens 2 is mounted within the farther boxed region, at distance d. Each problem in the table refers to a different combination of lenses and different values for distances, which are given in centimeters. The type of lens is indicated by C for converging and D for diverging; the number after C or D is the distance between a lens and either of its focal points (the proper sign of the focal distance is not indicated). Find (a) the image distance i2 for the image produced by lens 2 (the final image produced by the system) and (b) the overall lateral magnification M for the system, including signs.

Explanation / Answer

a) Ok, for this lens system, we need first to work with the first lens, then this lens will make an image, we will use this image as an object for the other lens, and we will find the final location of the image,

Note, the lens are 13 cm apart each other, and the object is at 20 cm to the left of the first lens

For the first lens:

1 / p + 1 / q = 1 / f

p = object distance

q = image distance

1 / 20 + 1 / q = - 1 / 21

q = -10.24 cm

So the first image is 10.24 cm to the left of the first lens, and this image is at (13+10.24 = 23.24) cm to the left of the second lens, now, let's work with the image and the second lens :

1 / 23.24 + 1 / q = -1 / 9.8

q = -6.9 cm

So, well, the final image will be 6.9 cm to the left of the second lens.

b) Since the distance is negative with respect to the final lens, the image is virtual. As per the magnification, we have that the total magnification is give by

Magnification = (12.5/50) * ( 0.454/0.5) = 0.227

Overall magnification

M = (q1/p1) / (q2/p2) = 10.24*23.24 / 20*6.9 = 1.72

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