Two-lens systems. In the figure, stick figure O (the object) stands on the commo
ID: 1493650 • Letter: T
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. Also, determine whether the final image is (c) real or virtual, (d) inverted from object O or noninverted, and (e) on the same side of lens 2 as object O or on the opposite side.
(a) (b) (c) (d) (e) p1 Lens 1 d Lens 2 i2 M R/V I/NI Side +13 C, 8.8 31 D, 9.2
Explanation / Answer
We shall us the cartesian coordinate system where the lens is placed at the origin and so, the distances measured to the right are taken positive and those measured to the left are taken to be negative. Alo, distances measured upwards are taken positive and those measured downwards are taken to be negative.
In this cartesian system, the incident light ray always comes from the left and the focal length of a Converging lens is +ve and that of a diverging lens is - ve.
focal length of the first lens is, therefore f1= 8.8 cm
and object distance u = - 13 cm (-ve because it is to the left of the lens)
using the lens formula 1/f = 1/v - 1/u
1/8.8 = 1/v - 1/-13 => 1/v = 1/8.8 - 1/13 = 0.036713286
v = 27.23809524 cm to the right of the first lens.
So, it is at a distance of 31 - 27.23809524 = 3.761904762 cm to the left of the second lens
For the refraction through the second lens, this image acts as the object. So, object distance for the second lens = u2 = - 3.76190472 cm (-ve because it is to the left of the second lens)
f2 = - 9.2 cm
using the lens formula 1/f2 = 1/v2 - 1/u2
1/ -9.2 = 1/v2 - 1/ -3.76190472
1/v2 = - 0.374518437
v2 = - 2.67 cm
negative sign indicates that the final image is formed to the left of the second lens.
magnification of first lens m1 = v1/u1 = 27.23809524 / - 13 = -2.095238
magnification of second lens m2 = v2/u2 = -2.67 / - 3.76190472 = 0.709746843
Combined magnification M = m1xm2 = -2.095238 x 0.709746843 = -1.487088557
The first lens gave a real and inverted image as output. This acted as object for the second lens, where the divinging lens formed a virtual image, (to the same side, which means the rays didn't intersect to form a real image). So, the final image is virtual and the inverted image (by the first lens) remains inverted. And as we have seen, the final image is to the lefgt of the second lens.
Answers: a.) - 2.67 cm b.) -1.487 c.) virtual d.) inverted e.) left side of the second lens
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