In the figure, an object is placed in front of a converging lens at a distance e

Question

In the figure, an object is placed in front of a converging lens at a distance equal to twice the focal length f_1 of the lens. On the other side of the lens is a concave mirror of focal length f_2 separated from the lens by a distance 2(f_1 +f_2). 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_1 = 7.6 cm and f_2 = 6.4 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?

Explanation / Answer

The image from the lens is

1/f1 = 1/do + 1/di =>

1/di = 1/f1 - 1/do = 1/f1 - 1/(2f1) = 1/(2f1)

so di = 2f1 = 2*7.6 = 15.2cm

The distance of this image from the mirror = 2*(f1 + f2) - 15.2 = 2*(7.6 + 6.4) - 15.2 = 12.8cm

Now this is the object distance for the mirror

So the image from the mirror is

1/f2 = 1/do + 1/di => 1/di = 1/f2 - 1/do

or di = f2*do/(do - f2) = 6.4*12.8/(12.8 - 6.4) = 12.8cm

Now this means the object for the lens is 2(f1 + f2) -12.8 = 15.2cm

Finally a) 1/f1 = 1/do + 1/di or 1/di = 1/f1 - 1/do

=> di = f1*do/(do - f1) = 7.6*15.2/(15.2 - 7.6) = 15.2cm to the left of the lens

b) M = m1*m2*m3

m1 = -di1/do1 = -15.2/15.2 = -1

m2 = -di2/do2 = -12.8/12.8 = -1

m3 = -di3/do3 = -15.2/15.2 = -1

So M = (-1)*(-1)(-1) = -1

Real and inverted

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