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course PHY 232
11:09 am 3/17
110302 Physics II`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 proportionality.
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These geometric problems are hard to describe well. There are 2 sets of 2 similar triangles. The first set relates the two image/object heights to the object/image distances, i.e. o/y= i/y'. Then the other set has a relation such that y/f = y'/(i-f).
y=i*y'/o
o*y'/i = fy'/(i-f)
o/i = f/(i-f)
o= fi/(i-f)
oi-of = fi
i = o(i-f)/f
This can also be rearranged to the very simple form:
1/f = 1/i+1/o
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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?
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?
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.
Challenge question: Two lenses of focal lengths f_1 and f_2 are to be used to form an image of an overhead light, with the image being formed on a screen at distance s from the light. If the first lens is held at distance s_1 below the light, then for what values of s_1 is it possible, using the second lens, to form on the screen an image of the first image? In terms of s_1, what is the distance s_2 of the second lens from the light?
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1 / i = 1 / f + 1 / o
I don't see how that expression is right since it should be what I derived above, in which case:
i = 10.9 cm
i_1 = o_2= 20, f_2 = 10 cm
i_2 = 20 cm
It is possible if you fix the second lens at o_2= 15 cm, f_2 = 10 cm then i_2 = 30 cm. This adds to make 45 cm so you can subtract that from 180 to get 135 cm. Now I used a little trial and error and found that o_1 = 124 makes i_1 = 10.88 cm which adds to get very close to 180 and you could solve for exactly 180 using the mess_1 and mess_2 expressions.
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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?
Give three film thicknesses that will cause positive interference, and three that will cause negative interference.
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2*(450)/600 = 3/2 which is 1/2 lambda OOP so they have complete negative interference. Positive means whole number multiples of lambda so: 300, 600, 1200 and negative means anything else really but complete negative interference occurs when whole number + 1/2 multiples lambda are present: 150, 750, 1050.
@& Good. Of course the index of refraction of the oil has an effect, but I let the problem stand in its oversimplified form.
The index of refraction results in a shorter wavelength in the oil. For index 1.35, an oil film of thickness 450 nm / 1.5 = 300 nm would have the calculated effect.*@
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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?
If you walk at .5 meter / second in the direction opposite that of the elephants, with what frequency will you need to smile?
If you back up at .5 meter / second, with what frequency will you need to smile?
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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2 m/s * 1/5 sm/m = 0.4 sm/s
Your relative velocity to the elephants is 2 + 0.5 = 2.5 m/s since the elephants are now appearing more frequently so: 2.5 m/s * 1/5 sm/m = 0.5 sm/s
Now your relative velocity is 2-0.5= 1.5 m/s since elephant frequency is less: 1.5 m/s * 1/5 sm/s = 03/ sm/s
If the elephants walk at 2 m/s on a track that is moving 0.5 m/s the net velocity is additive again and is 2.5 m/s thus 0.5 sm/s
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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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A pattern of ANA that AN= 108/2 = 54 --> 54*4 = 216 cm = 2.16 m. f = v/lambda = 6400/2.16 = 2963 Hz
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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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15 m/ 340 m/s = 0.044s
But, from the elephant scenario we know the rod added, or subtracted, from the net velocity so that it would be 340±v_rod.
@& It's not quite as simple as the elephant scenario. Relative velocity isn't the key here, since the wave is propagating through a stationary medium.*@
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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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2500 cycles/s * 1/340 s/m = 7.35 c/m --> 7.35*12 = 88.2 cycles --> 88.2 cycles/ 2500 cycles/s = 0.035s
t = 0.035 s
# peaks = 88.2
I'm not sure about this because this means the rod was moving at 340 m/s and it certainly wasn't that fast.
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How long did it take first peak to reach the microphone?
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12/340 = 0.035s
This result confirms that something is wrong above since this means that both the rod and the first peak hit the mic at the same time.
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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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This seems like a more reasonable speed.
12m/ 20 m/s = 0.6 s
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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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0.6-0.035 =0.565 s
But again it seems that the speed of the rod affects the sound speed to make it 320 m/s relative to the mic.
So the wave actually hit at 12/320 = 0.0375--> 0.563s
@& The speed of the rod does not affect the speed with which the sound propagates.*@
@& Your first result was correct.*@
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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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2500*0.6 =1500 cycles occurred in 0.563s
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What therefore was the frequency detected by the microphone?
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f = 1500/0.563 = 2664 Hz
The books formula gives 2656 Hz, the difference is probably due to rounding but I've spent way too much time trying to reason out how the velocities add/subtract. It makes sense that the source moving toward you increases frequency but that the velocities subtract is harder for me because relative to the listener the speed of the wave is 360 m/s but relative to the rod it's 320 m/s and I guess it's the v_rod that is important and not the v_listener. It's hard for me to just accept this but it seems true.
@& The key is that the speed of the rod doesn't affect the speed of the sound. Eliminate that assumption and your result will agree with that of the book's formula.*@
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`q007. Suppose that rather than me throwing the rod at you, you threw the microphone at the rod at 20 m/s, while I held it stationary. Use an analysis similar to the one above to find the frequency that would be detected by the microphone. Your answer will be close to, but not identical with, the answer you got above.
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Here is where the velocities seem to add which is how I kept trying to work the previous problem. The relative increase in v= 1.0588 whereas the relative v above was 1.0625 which is why the results are similar but not exact since it's the difference between 340/320 and 360/340. Anyway:
f = 2647 Hz
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`q008. If there was a microphone stationary at the rod, and its signal was mixed with the signal from the microphone you threw, how many beats per second would be observed?
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f_beats= 2647-2500 = 147 Hz
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`q009. From the measurements you made in Monday's class, what do you get for the wavelength of laser light?
**** I missed this day and thus do not have measurements for this.
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@& Remind me and we can set it up in about 2 minutes.*@
&#Good work. See my notes and let me know if you have questions. &#