A diverging lens (f = -20.0 cm) is placed 10.0 cm in front of a plane mirror, A

Question

A diverging lens (f = -20.0 cm) is placed 10.0 cm in front of a plane mirror, A matchstick is placed 25.0 cm in front of the lens.

(a) Calculate the location of the final image seen by someone who looks through the lens toward the mirror. Explain your reasoning in each step. (hint: The light from the object travels through the lens, reflects off the mirror, goes back through the lens, and into the eyes. The image formed by the lens is the object for the mirror. The image formed by the mirror is the object for the lens the second time the light goes through it. The mirror reverses the direction of the light, and also the meaning of the front and back for the lens.)

(b) Is the final image real or virtual? Explain.

(c) What is the overall magnification of the image?

(d) Is the image upright or inverted, compared to the object? Explain how you know.

Explanation / Answer

Given,

f1 = -20 cm ; o1 = 25 cm ; d = 10 cm ; f2 = infinity

let i1 be the image distance formed by lens.

from lens equation, 1/f = 1/i + 1/o

i = o x f / (o - f) =

i1 = o1 x f1 / (o1 - f1) = 25 x -20 / (25 + 20) = 11.11 cm

o2 = 11.11 - 10 = 1.11 cm

again using lens equation to find the image distance i2 by plane mirror

1/infinity = 1/i2 + 1/o2

i2 = -o2 = -1.11 cm

i3 = o2 x f1 / (o2 - f1) = -1.11 x -20 / ( -1.11 + 20) = 1.18 cm

Hence, the location of the final image is i3 = 1.18 cm.

(b)Final image is real.

m1 = -i1/o1 = -11.11/25 = -0.56

m2 = -i2/o2 = -1.11/1.11 = -1

m3 = -i3/o2 = -1.18/(-1.11) = 1.06

The oveall magnification is 1.06.

(b) The image is upright. since the magnification is positive.

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