Asst 30 Polarization

course phy 202

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030. `Query 19

Physics II

09-06-2007

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19:45:42

Principles 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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RESPONSE -->

tan theta p = n

theta p = tan ^-1 (1.52)

theta p = 1.52 degrees

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19:46:22

Brewster'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

theta_p = 56.7 degrees, approximately.

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RESPONSE -->

I had that answer, I just typed in the wrong number.

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19:53:19

gen phy problem 24.43 foil separates one end of two stacked glass plates; 28 lines observed for normal 650 nm light

gen phy what is the thickness of the foil?

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RESPONSE -->

2t = m *lambda

t = 28 (650 * 10^-9) / 2

t = 9.1 * 10^-6 m

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19:59:34

STUDENT 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(2)(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) **

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RESPONSE -->

I think I got the correct answer, though I used the numbers you gave rather than those in the book.

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20:03:44

**** gen phy how many wavelengths comprise the thickness of the foil?

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RESPONSE -->

9.1c* 10^-6 m / 6.5 *10^-9 m = 1400 wavelengths

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20:11:08

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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RESPONSE -->

I used the wrong method. I should have done:

2t = m lambda

2 (9.1*10^-6) / (6.5*10^-9) = 2800

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Your work looks good. Let me know if you have any questions. &#