course PHY 232
July 12 at 5:30
Self-critique Rating:*********************************************
Question: `q**** query univ phy problem 35.52 (37.46 10th edition) normal 477.0 nm light reflects from glass plate (n=1.52) and interferes constructively; next such wavelength is 540.6 nm.
How thick is the plate?
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Your solution:
I am having trouble picturing what this situation looks like. If I cant picture a problem I have a lot of trouble solving it.
Given Solution:
`a** The path difference for constructive interference is an integer multiple of the wavelength. The path difference here is twice the thickness.
Wavelengths in glass are 477 nm / 1.52 and 540.6 nm / 1.52.
So we know that double the thickness is an integer multiple of 477 nm / 1.52, and also an integer multiple of 540.6 nm / 1.52.
We need to find the first integer multiple of 477 nm / 1.52 that is also an integer multiple of 540.6 nm / 1.52.
We first find an integer multiply of 477 that is also an integer multiply of 540.6.
Integer multiples of 540.6 are 540.6, 1081.2, 1621.8, etc. Dividing these numbers by 477 we obtain remainders 63.6, 127.2, etc. When the remainder is a multiple of 477 then we have an integer multiple of 477 which is also an integer multiple of 540.6.
SInce 477 / 63.6 = 8.5, we see that 2 * 477 / 63.6 = 17. So 17 wavelengths of 477 cm light is the first multiple that is equivalent to an integer number of wavelengths of 540.6 cm light.
17 * 477 = 8109.
Since 8109 / 540.6 = 15, we see that 17 wavelengths of 477 nm light span the same distance as 15 wavelengths of 540.6 nm light.
It easily follows that that 17 wavelengths of (477 nm / 1.52) light span the same distance as 15 wavelengths of (540.6 nm / 1.52) light.
This distance is 17 * 477 nm / 1.52 = 5335 nm.
This is double the thickness of the pane. The thickness is therefore
pane thickness = 5335 nm / 2 = 2667 nm.
IF INTERFERENCE WAS DESTRUCTIVE: n * 477 nm / 1.52 = (n-1) * 540.6 nm / 1.52, which we solve:
Multiplying by 1.52 / nm we get
477 n = 540.6 n - 540.6
n * (540.6 - 477 ) = 540.6
n * 63.6 = 540.6
n = 540.6 / 63.6 = 8.5.
This is a integer plus a half integer of wavelengths, which would result in destructive interference for both waves.
Multiplying 8.5 wavelengths by 477 nm / 1.52 we get round-trip distance 2667 nm, or thickness 1334 nm. **
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Self-critique (if necessary):
Self-critique Rating:
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Question: `q**** query univ phy prob 35.50 (10th edition 37.44): 700 nm red light thru 2 slits; monochromatic visible ligth unknown wavelength. Center of m = 3 fringe pure red. Possible wavelengths? Need to know slit spacing to answer?
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Your solution:
For this problem we know the wavelength of the red light, that the red light and the monochromatic light passes through a two slit apparatus, and that at m=3, the two combined are bright red.
We want to find the wavelength of the second light for which this could happen.
If both of the lights were in constructive interference at this point, the light would probably not be red because both would show and form some other color.. So I think that the two are destructive.
So, d*sin(theta) = (4/3)*lambda
I am unsure how where to go from here.
It seems that you would need the slit distance in this case.
Given Solution:
`aSTUDENT SOLUTION: The pure red band at m = 3 suggests that there exists interference between the wavelength of the red light and that of the other light. Since only the red light is present at m = 3 it stands to reason that the wavelength of the other light is a half of a wavelength behind the red wavelength so that when the wavelength of the red light is at its peak, the wavelength of the other light is at its valley. In this way the amplitude of the red light is at its maximum and the amplitude of the other light is at it minimum – this explains why only the red light is exhibited in m = 3.
INSTRUCTOR COMMENT
At this point you've got it.
At the position of the m=3 maximum for the red light the red light from the further slit travels 3 wavelengths further than the light from the closer. The light of the unknown color travels 3.5 wavelengths further. So the unknown wavelength is 3/3.5 times that of the red, or 600 nm.
You don't need to know slit separation or distance (we're assuming that the distance is very large compared with the wavelength, a reasonable assumption for any distance we can actually see. **
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Self-critique (if necessary):
???? I was wondering in this problem whether both lights were coming through both slits or if one of the slits has only red light going through it, and the other slit has the monochromatic light going through it ????
The full problem statement is in your text. Your first alternative is the correct one; both light sources come through both slits.
Self-critique Rating:
See my notes on that second problem.
The text describes and shows diagrams of systems similar in nature to the one in the first problem.
&#Let me know if you have questions. &#