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course Phy 242
RESUBMITTING this, never showed up in my portfolio.
110302`q001. Light parallel to the axis of a thin lens is focused at distance f from the lens. A parallel beam originates at the tip of a candle at distance o from the lens and is refracted through the focal point. Another beam originates from the same point, strikes the lens at its center and passes through the lens undeflected.
At what distance i from the lens do the two rays meet? Answer by sketching the rays, forming a series of similar triangles and solving the resulting proportionalities.
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i =[(i*f)*o] / f
(1/i) + (1/o) = (1/f)
(i/FE) = [(i-f)*o] / (f*FE)
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`q002. The image of an object originally at distance o from a lens with focal length f forms at distance i from the lens, where 1 / i = 1 / f + 1 / o.
If f = 10 cm, at what distance from the lens will the image of an overhead light 120 cm above the lens be formed?
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(1/10) = (1/i) + (1/120)
0.10cm^(-1) = (1/i) + 0.008cm^(-1)
(1/i) = 0.092cm^(-1)
I = (1/0.092cm^-1) = 11cm approx
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If another lens of equal focal length is placed 20 cm below that image, it will form an image of the image. How far below the lens will the image of the image form?
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o = 20cm and i =10cm
(1/10) = (1/i) + (1/20)
0.05 = (1/i)
I = 20cm
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Is it possible to form an image of the first image on a tabletop 180 cm below the lens? If not, show why not. If so, show how far below the first image the lens must be placed.
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No
@&
As shown in the Class Notes, f = 10 cm and o = 120 cm yield the equation
1 / (10 cm) = 1 / i + 1 / (120 cm)
with solution
i = 11 cm (approx..)
Another lens of equal focal length, 20 cm below that image, will 'see' that image as its object. So for that lens
o = 20 cm
and we have
1 / (10 cm) = 1 / i + 1 / (20 cm)
with solution
i = 20 cm.
From the overhead light to the image the distance is the sum of 120 cm (light to first lens), 11 cm (first lens to first image), 20 cm (first image to second lens) and 20 cm ( second lens to second image).
This is a total distance of 171 cm.
If we hold a screen 171 cm below the light, with the two lenses positioned accordingly, and shield the second lens from the direct light of the overhead source, we should see the second image.
By adjusting the two lenses slightly, it should be possible to form an image on the tabletop. It is possible to form the image on the tabletop with the first lens in the position specified:
The tabletop is 49 cm below the image formed by the first lens. If an image is to be formed on the tabletop, then, the distances i and o must add up to 49 cm, so o = 49 cm - i and we have the equation
1 / 10 cm = 1 / i + 1 / (49 cm - i)
This equation is not difficult to solve. The result is i = 14 cm approx., so o = 49 cm - 14 cm = 35 cm.
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`q003. Light of wavelength 600 nm strikes a thin film of oil on top of a layer of water. Some of the light reflects off the surface of the film, and some off the surface of the water.
If the thickness of the film is 450 nm, will the two reflected beams interfere positively, negatively, or somewhere in between?
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Negatively
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Give three film thicknesses that will cause positive interference, and three that will cause negative interference.
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The wavelength of the light as is passes through the oil is in fact less than the 600 nm wavelength. If the index of refraction of air is very close to 1 and the index of refraction of the oil is 1.5, then how does this affect your results?
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@&
The light that passes through the oil layer will travel 450 nm to the water surface, from which some of the light reflects and travels another 450 nm back through the oil, a total of 900 nm.
For simplicity we first assume that the wavelength of the light doesn't change when it moves from air into oil. It does, and we'll account for that shortly. However if it doesn't then:
If the wavelength of the light remains 600 nm, then, the light reflecting off the water will travel 900 nm / (600 nm / wavelength) = 1.5 wavelengths further than light reflecting off the oil surface. This will result in destructive interference, with the peaks of one wave meeting the valleys of the other.
Positive interference would result if the path difference is 1, 2, 3, 4, ... wavelengths, i.e., with current assumptions, if path difference is 600 nm, 1200 nm, 1800 nm, ... . The path difference is twice the thickness of the oil layer, so the layers of thickness 300 nm, 600 nm, 900 nm, ... would result in positive interference.
Path differences of 1/2, 3/2, 5/2, 7/2, ..., wavelengths result in destructive interference, which would therefore occur for layers of thickness 150 nm, 450 nm (as we have seen), 750 nm, ... .
Now in fact oil with index of refraction 1.5 will decrease the speed of the light by a factor of 1.5. The frequency of the light remains unchanged, so its wavelength is reduced by factor 1.5 to 400 nm. So with oil of this refractive index, the 900 cm path difference corresponds to 900 nm / (400 nm/wavelength) = 2.25 wavelengths. This is between 2 wavelengths, which would result in constructive interference, and 2.5 wavelengths, which would result in destructive interference.
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`q004. A long line of elephants, trunk to tail, is passing you. The elephants move at a steady speed of 2 meters / second, and each elephant is 5 meters long. Your job is to smile in greeting to every elephant that passes you. With what frequency do you have to smile?
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Every 2.5 seconds
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If you walk at .5 meter / second in the direction opposite that of the elephants, with what frequency will you need to smile?
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Every other second, more frequently
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If you back up at .5 meter / second, with what frequency will you need to smile?
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Every other second
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If the elephants are on a moving walkway, which is moving at .5 meter / second in your direction, with you standing, with what frequency will you need to smile?
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Every second
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`q005. The speed of sound in aluminum is about 6400 m/s. The aluminum rod we used was about 108 cm long. I held the rod in the middle, creating a node at that point. The ends of the rod were free to vibrate so there are antinodes at the ends. What therefore was the frequency of the sound created in that rod?
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3200 pulse/sec
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`q006. When I threw the 'singing rod' at you, at the instant it left my hand it was about 15 meters away. The first 'peak' of the sound wave that left the end of the rod after I released it was 15 meters from you and hadn't yet reached your ear.
Moving at 340 m/s, how long did it take that first peak to get to your ear?
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Wouldn’t you use the “Doppler effect” to solve this?
@& It would, and you did it right on the test.
However the reasoning behind the formula can be understood in terms of these examples.
The first peak moves 15 meters at 340 m/s, arriving at your ear about .044 seconds, approximately, after the rod is released.
It reaches to the microphone in about .035 seconds after release.
The rod, on the other hand, moved at 20 m/s and therefore required .60 seconds to reach the microphone. In that time, at 2500 Hz, it produced 2500 cycles / sec * .6 sec = 1500 cycles.
The last cycle was produced at the microphone, .6 sec - .035 sec = .565 sec after the first cycle was received by the microphone.
So the microphone received 1500 cycles in .565 seconds. The frequency of the sound received was therefore 1500 cycles / (.565 sec) = 2650 cycles / sec, approx..
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The rod didn't hit you in the ear. I threw it so that it would land a few meters in front of you, so as to avoid damaging the rod (or you). Suppose that there was a microphone at the point where the rod hit, so that the rod actually hit the microphone at a distance 12 meters from where I released it, and suppose that you were listening to the signal from the microphone
You will agree that in traveling that 12 meters the rod emitted a number of peaks. How long did it take the rod to get to that point, and how many peaks were produced, if we assume that the frequency of the sound produced is 2500 Hz?
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12m = (velocity/2500Hz)
Velocity = 30 000 m/s
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How long did it take first peak to reach the microphone?
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15 000 m/s
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Assuming that the rod traveled at 20 meters / second, how long did it take the rod to reach the microphone?
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C = f(lambda)
C = (2500Hz)*(20m/s)
= 50 000m
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From the time the first peak reached the microphone, until the rod hit the microphone, how much time elapsed?
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How many peaks were therefore detected by the microphone, and during what time interval were they detected?
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What therefore was the frequency detected by the microphone?
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The microphone would have reached the rod in the same .6 seconds it took the rod to reach the microphone (this since the assumed speeds of the two throws were equal).
The first peak received by the microphone was produced by the rod .035 seconds earlier prior to the throw.
So all the peaks produced in the .035 seconds prior to the throw, and the .60 seconds required for the microphone to reach the rod, were received in .60 seconds.
In .635 seconds the number of peaks produced was about 1590. So about 1590 peaks were received in .60 seconds, resulting in a frequency of 1590 cycles / .60 sec = 2650 Hz.
This appears to agree with the result of the preceding, but if the calculations are done accurately there will be a small but significant difference in the result.
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&#Good responses. See my notes and let me know if you have questions. &#