The figure below shows a nickel and its image. The image has 2.2 times the diame

Question

The figure below shows a nickel and its image. The image has 2.2 times the diameter of the nickel and is 3.07 from the lens (i.e. the magnification factor is 2.2)

a) Is the image real or virtual

b) Determine the focal length of thr lens.

c) Suppose you now use this lens to read a news paper but your near point is 15cm. how far from the newspaper should the lens be?

d) Now suppose that you use this lens as part of a camera. you want to photograph an object which is 30cm to the left of the lens. How far to the right of the lens should the film be?

The figure below shows a nickel and its image. The image has 2.2 times the diameter of the nickel and is 3.07 from the lens (i.e. the magnification factor is 2.2) Is the image real or virtual Determine the focal length of thr lens. Suppose you now use this lens to read a news paper but your near point is 15cm. how far from the newspaper should the lens be? Now suppose that you use this lens as part of a camera. you want to photograph an object which is 30cm to the left of the lens. How far to the right of the lens should the film be?

Explanation / Answer

the image is a virtual image ... because if the image were real it would have been inverted .. but in the picture it is upright

all magnifying glasses work in this way only

image distance = -3.07 cm

magnification = - image distance / object distance

so object distance = 1.4

using the lens equation

1/f = 1/i + 1/o

1/f = 1/(-3.07) + 1/1.4

f=2.574 cm --------------- focal length

supposing that we read an upright news paper ... the lens can be place anywhere between the 2.574 cm to 0 cm ...

but 0 cm is not practical ..

object distance = 30cm

image distance = ?

1/2.574= 1/i + 1/30

i = 2.816 cm to the right of the lens

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