1. A certain LCD projector contains a single thin lens. An object 24.2 mm high i
ID: 3163747 • Letter: 1
Question
1. A certain LCD projector contains a single thin lens. An object 24.2 mm high is to be projected so that its image fills a screen 1.80 m high. The object-to-screen distance is 3.00 m. Determine the focal length of the projection lens.Hints: You can determine if the image is real or virtual. Note that the object to screen distance is not p. Be careful to put all distances in the same units, but you can use either mm or m everywhere. The correct answer is in the range of: 32.1 mm to 45.5 mm
2. A diverging lens is to be used to produce a virtual image 0.329 times as tall as the object. Find where should the object be placed and then enter | p/f |.
Hints: The correct answer is in the range of: 1.20 to 3.55
1. A certain LCD projector contains a single thin lens. An object 24.2 mm high is to be projected so that its image fills a screen 1.80 m high. The object-to-screen distance is 3.00 m. Determine the focal length of the projection lens.
Hints: You can determine if the image is real or virtual. Note that the object to screen distance is not p. Be careful to put all distances in the same units, but you can use either mm or m everywhere. The correct answer is in the range of: 32.1 mm to 45.5 mm
2. A diverging lens is to be used to produce a virtual image 0.329 times as tall as the object. Find where should the object be placed and then enter | p/f |.
Hints: The correct answer is in the range of: 1.20 to 3.55
1. A certain LCD projector contains a single thin lens. An object 24.2 mm high is to be projected so that its image fills a screen 1.80 m high. The object-to-screen distance is 3.00 m. Determine the focal length of the projection lens.
Hints: You can determine if the image is real or virtual. Note that the object to screen distance is not p. Be careful to put all distances in the same units, but you can use either mm or m everywhere. The correct answer is in the range of: 32.1 mm to 45.5 mm
2. A diverging lens is to be used to produce a virtual image 0.329 times as tall as the object. Find where should the object be placed and then enter | p/f |.
Hints: The correct answer is in the range of: 1.20 to 3.55
2. A diverging lens is to be used to produce a virtual image 0.329 times as tall as the object. Find where should the object be placed and then enter | p/f |.
Hints: The correct answer is in the range of: 1.20 to 3.55
Explanation / Answer
Here,
ho = 24.2 mm
hi = 1.80 m
di + do = 3 m ----(1)
as di/do = hi/ho
di/do = 1.8/.0242 ----(2)
solving for di and do
do = 0.0397 m
di = 2.96 m
also ,
1/f = 1/di + 1/do
1/f = 1/.0397 + 1/2.96
solving for f
f = 0.0392 m = 3.92 cm
the focal length of lens needed is 3.92 cm = 39.2 mm
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