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

Question

3. 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. (Check your answer: 2.2 cm behind the mirror.) Be sure to show all your work, and 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 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

for diverging lens -1/f =1/u + 1/v

so here

-1/20    = 1/25 + 1/v

1/v = -1/20 - 1/25

v = -11.11 cm

for plane mirror, u = -10-11.11 = -21.11 cm

object distance = 21.11 cm

for concave mirror u = -21.11 -10 = -31.11 cm

new object distance u = 31.11 cm

1/f = 1/31.11 +1 /v

1/v = -1/20 - 1/31.11

v   = -12.2 cm

location of final image = 12.2 - 10 = 2.2 cm (ANSWER)

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b) The image formed is virtual because of production of virtual images by concave lens then mirror and then again lens.

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c) Magnification(1) = 11.11/25 = 0.444

Height of Image(1) = 0.444 * Height of object

Magnification from mirror =1

Magnification(2) = 12.17/31.11

M2 = 0.391

Height of final Image = 0.391* H1

Hf = 0.391*0.444

Height of object = 0.174* Height of object

So, overall magnification = 0.174
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d) The final image is upright as shown

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