You have a small but bright computer display that is only 4 cm wide by 4 cm tall

Question

You have a small but bright computer display that is only 4 cm wide by 4 cm tall. You will place a lens near the display to project an image of the display about 2.1 m tall onto a large screen that is 9 m from the lens of the projector.

(a) Should you use a converging lens or a diverging lens? converging

(b) What should the focal length of the lens be? (Give your answer to at least two decimal places.)

(c) Suppose that you choose a lens that has the focal length that you calculated in part (b). What exactly should be the distance from the computer display to the lens? (Give your answer to at least two decimal places.) 17.1429 Correct: Your answer is correct.

(d) In order that the screen display be right side up, should the computer display be inverted or right side up? inverted Correct: Your answer is correct.

(e) On another occasion the screen is moved close, so that it is only 5 m from the lens. To focus the image, you have to readjust the distance between the computer display and the lens. What is the new exact distance from the computer display to the lens? (Give your answer to at least two decimal places.)

(f) How tall is the image on the screen, now that the screen is only 5 m from the lens?

Explanation / Answer

Height of object, ho = 4 cm
Height of image, hi = 2.1 m
Distance of image, di = 9 m

M = hi/ho = (2.1*100)/4
M = 52.5
M = di/do
do = 9/52.5 = 0.171428 m
do = 17.1429 cm

(b)
Focal length of the lens ,
1/f = 1/di + 1/do
1/f = 1/9 + 1/0.171428
f = 0.168 m
f = 16.8 cm

(c)
As calculated above,
do = 17.1429 cm

(d)
Computer display should be inverted.

(e)
Now,
Distance of image = 5 m
f = 16.8 cm
1/f = 1/di + 1/do
1/16.8 = 1/500 + 1/do
do = 17.384 cm
new exact distance from the computer display to the lens, do = 17.384 cm
(f)
hi/ho = -di/do
hi = (500/17.384) * 4cm
hi = 115.0 cm

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