- #1
mathzing
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hey, i tried the following problem and i think i got it wrong well. here is the problem and what i did.
problem
A photograph of a zebra is taken from a distance of 100m with a 50mm lens. A second picture is taken from a distance of 50m with the same lens. What is the ratio of the height of the image in the second picture to the height in the first picture?
What i did
well. its obvious that the ratio is 2:1 because the distance of image int eh 100m to the distance of the 50m picture is 2:1 and the fact that it uses the same lens proves it. but my main issue with the problem is that i can't understand weather the 50mm lens can be used as the focal length of the problem. well I used it and here is what i got:
for the 100m:
1/f = 1/do + 1/di
so: 1/0.050 = 1/100 + 1/di
5di/0.050 = 5di/100 + 5di/di
100di = 0.05di + 5
100di - 0.05di = 5
99.95di = 5
99.95di/99.95 = 5/99.95
Therefore: di = 0.050025012
Note: to get the height of the image or object the equation is:
M = hi/ho = -di/do
I have the di and do but I don’t understand how I get hi or ho
M is the magnification For the 50m:
1/f = 1/do + 1/di
So: 1/0.050 = 1/50 + 1/di
2.5di/0.050 = 2.5di/50 + 2.5di/di
50di = 0.050di + 2.5
50di – 0.050di = 2.5
49.95di = 2.5
49.95di/49.95 = 2.5/49.95
Therefore: di = 0.05005005
..here is the confusing part: the di of the 100 its half of the di or the 50 which I think it should be, and I don’t see how I can get the highet of the image.?
problem
A photograph of a zebra is taken from a distance of 100m with a 50mm lens. A second picture is taken from a distance of 50m with the same lens. What is the ratio of the height of the image in the second picture to the height in the first picture?
What i did
well. its obvious that the ratio is 2:1 because the distance of image int eh 100m to the distance of the 50m picture is 2:1 and the fact that it uses the same lens proves it. but my main issue with the problem is that i can't understand weather the 50mm lens can be used as the focal length of the problem. well I used it and here is what i got:
for the 100m:
1/f = 1/do + 1/di
so: 1/0.050 = 1/100 + 1/di
5di/0.050 = 5di/100 + 5di/di
100di = 0.05di + 5
100di - 0.05di = 5
99.95di = 5
99.95di/99.95 = 5/99.95
Therefore: di = 0.050025012
Note: to get the height of the image or object the equation is:
M = hi/ho = -di/do
I have the di and do but I don’t understand how I get hi or ho
M is the magnification For the 50m:
1/f = 1/do + 1/di
So: 1/0.050 = 1/50 + 1/di
2.5di/0.050 = 2.5di/50 + 2.5di/di
50di = 0.050di + 2.5
50di – 0.050di = 2.5
49.95di = 2.5
49.95di/49.95 = 2.5/49.95
Therefore: di = 0.05005005
..here is the confusing part: the di of the 100 its half of the di or the 50 which I think it should be, and I don’t see how I can get the highet of the image.?
