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course PHY 202
3/11/15 at 5:45PM
021. `Query 19
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Question: `qPrinciples of Physics and General College Physics Problem 24.54: What is Brewster's angle for an air-glass interface (n = 1.52 for glass)?
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Your solution:
Tan(theta) = n2/n1
Tan(theta) = 1.52/1.0
Theta = 56.7 degrees
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Given Solution:
`aBrewster's angle is the smallest angle theta_p of incidence at which light is completely polarized. This occurs when tan(theta_p) = n2 / n1, where n2 is the index of refraction on the 'other side' of the interface.
For an air-glass interface, n1 = 1 so tan( theta_p) = n2 / 1 = n2, the index of refraction of the glass. We get
tan(theta_p) = 1.52 so that
theta_p = arcTan(1.52). This is calculated as the inverse tangent of 1.52, using the 2d function-tan combination on a calculator. WE obtain
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There would be no need to use radians on this problem.
However the small-angle approximations
sin(theta) = tan(theta) = theta
hold only if theta is in radians.
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theta_p = 56.7 degrees, approximately.
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Self-critique (if necessary):
Thank you for the explanation on using radians on the calculator when calculating sines of small angles (in the question form I sent you). In this case I used degrees when making my calculation because using radians got me an answer smaller than 1, which I thought was not reasonable. So, do we use radians when the value is particularly small? I understand your explanation, which was very helpful, but I’m still a little unclear as to when we use radians versus degrees, because clearly in this problem we had to use degrees.
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Question: `qgen phy problem 24.44 foil separates one end of two stacked glass plates; 28 lines observed for normal 670 nm light
gen phy what is the thickness of the foil?
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Your solution:
Dark bands occur at 2t=m*lambda where m is an integer
2t = m*lambda
2t = (28)(670*10^-9m)
t = 9.38*10^-6m
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Given Solution:
`aSTUDENT SOLUTION: To solve this problem, I refer to fig. 24-31 in the text book as the problem stated. To determine the thickness of the foil, I considered the foil to be an air gap. I am not sure that this is correct. Therefore, I used the equation 2t=m'lambda, m=(0,1,2,...). THis is where the dark bands occur .
lambda is given in the problem as 670nm and m=27, because between 28 dark lines, there are 27 intervals.
Solve for t(thickness):
t=1/2(27)(670nm)
=9.05 *10^3nm=9.05 um
INSTRUCTOR RESPONSE WITH DIRECT-REASONING SOLUTION:** Your solution looks good. Direct reasoning:
** each half-wavelength of separation causes a dark band so there are 27 such intervals, therefore 27 half-wavelengths and the thickness is 27 * 1/2 * 670 nm = 9000 nm (approx) **
**** gen phy how many wavelengths comprise the thickness of the foil?
GOOD STUDENT SOLUTION: To calculate the number of wavelengths that comprise the thickness of the foil, I use the same equation as above 2t=m'lambda and solve for m.
2(9.05 um)=m(6.70 *10^-7m)
Convert all units to meters.
m=27 wavelengths.
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Self-critique (if necessary):
Hmm, I used 28 for m because there are 28 dark bands. The problem does state that there is one dark band at either end, though, so is this why we use 27 instead of 28?
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There is no interference at the end of the plate that lies on the table, so the first dark band occurs when there is a half-wavelength difference in the two paths.
Each of the 27 additional dark bands would correspond to another full wavelength of path difference.
By this reasoning, from the end on the tabletop to the last dark band the total path difference would be 27.5 wavelengths.
I don't have the text here, but if there is a dark band exactly an each end, then thethe path difference is only 27 wavelengths (same reasoning as fenceposts: 28 fenceposts implies 27 fence sections).
In practice you would analyze what you actually see. I would measure the distance from the first to the last dark line and find the corresponding path difference. I would then measure from the end of the glass pane on the tabletop to the foil, and multiply path difference by the ratio of the distances.
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Very good. Check my note(s).
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