Chapter 34, Problem 106 In the figure, an object is placed in front of a converg

Question

Chapter 34, Problem 106 In the figure, an object is placed in front of a converging lens at a distance equal to twice the focal length of the lens. On the other side of the lens is a concave mirror of focal length &separated; from the lens by a distance 2(f). Light from the object passes rightward through the lens, reflects from the mirror, passes leftward through the lens, and forms a final image of the object. Take f 8.1 cm and f 2.6 cm. What are (a) the distance between the lens and the final image and (b) the overall lateral magnification M of the object? Is the image (c) real or virtual (if it is virtual, it requires someone looking through the lens toward the mirror), (d) to the left or right of the lens, and (e) inverted or noninverted relative to the object? 0 (a) Number Units (b) Number Units

Explanation / Answer

Focal length of the lens = f1 = 8.1 cm

Object distance u = 2f1 = 16.2 cm

using Lens equation,

1/f = 1/v + 1/u

=> v = [(1/8.1) - (1/16.2)]-1 = 16.2 cm

this is the image distance from the lens [to the right].

distance between the lens and the mirror = D = 2[f1 + f2] = 21.40 cm

so, the object distance for the mirror will be:

u' = 21.40 - 16.20 = 5.20 cm

f2 = 2.6 cm

using the mirror equation:

1/2.6 = 1/5.2 + 1/v'

=> v' = 5.2 cm to the left of the mirror.

The object distance to the lens now will be:

u'' = 21.40 - 5.20 = 16.20 cm

f1 = 8.1 cm

therefore, v'' = 16.2 cm to the left of the lens.

so, the distance between the lens and the final image is: v'' = 16.2 cm.

b] Overall Magnification = M = - v''v'v/u''u'u = - 1.0.

c] The final image is real and inverted.

d] It is to the left of the lens.

e] It is inverted since the magnification is negative.

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