A lens with focal length 2.00 cm is placed 15.00 cm to the right of a mirror wit

Question

A lens with focal length 2.00 cm is placed 15.00 cm to the right of a mirror with focal length 6.00 cm (convex, reflective side facing to the right). An object is placed between the lens and the mirror, 6.00 cm from the lens. For an observer to the right of the setup, two images are formed, (I) from the light originally leaving the object traveling to the right, and (II) from the light originally leaving the object traveling to left. Using the analytical method (accurate ray tracing is not necessary), describe and comment on the similarities and differences between the following attributes of the two images. (a) Type – Real or virtual? (b) Orientation – Upright or inverted with respect to the original object? (c) Location – Calculate the distance between each of the images and the lens. Are the images to the right or the left of the lens? (d) Magnification – Calculate the total magnification of the two images.

Explanation / Answer

To answer the questions we find the position of each image

Part a ND B)

COMMENTARY in part c

Part c)

I) direct image through the lens

We use the equation builder

1/o + 1/i = 1/f

where

o is the distance to the object

i is the image distance

f is the focal length

Data

o = 6 cm

f = 2 cm

1/i = 1/f – 1/o

1/i = ½ – 1/6 = 0.5 - 0,1666

1/i = 0.33

i = 3 cm

to the right of the lens, the image is real and inverted

II) image formed by the mirror

data

o = 15-6 = 9 cm

f = -6.00 cm

1/o+ 1/i = 1/f

1/i = 1/f – 1/o

1/i = 1/(-6) -1/9 = - (0.166 +0.111)

1/i = 0.277

i = -3.6 cm

This image is formed by extensions of the rays that is formed behind the mirror and isa and right virtual image

Part D

for lens

m= - i/o

m = - 3/6 = -1/2

m = - 0.5

for the mirror

m= - (-3.6)/9

m = 0.4

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