A 5.00 cm tall object is 35.0 cm in front of a converging lens that has a 20.0 c

Question

A 5.00 cm tall object is 35.0 cm in front of a converging lens that has a 20.0 cm focal length. Use ray tracing to find the position and height of the image. Label the location of the object, s, the location of the image, s', and the image height, h', on your ray tracing diagram. Calculate the image positon, s', magnification, m, and image height, h'. Compare with your ray tracing answers in part a (small difference are acceptable). Is this image upright or inverted? Is this image real or virtual? How would the general characteristics (real/virtual, upright/inverted) change if the object were moved to 15.0 cm in front of the lens?

Explanation / Answer

a) the ray diagram you did was correct.

b) given data

focal length, f = 20 cm

object distance, s = 35 cm

let s' is image distance.

apply,

1/s + 1/s' = 1/f

1/s' = 1/f - 1/s

1/s' = 1/20 - 1/35

s' = 46.7 cm

magnification, m = -s'/s = -46.7/35 = -1.33

image height = m*object height

= 1.33*5

= 6.67 cm

c) image inverted

d) real

e) the image would be virtual and upright

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