course phy 122 assignment #030030. `Query 19
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09:03:12 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 --> Brewsters angle is the special angle of incidence that produces a 90 degree angle between the reflected and refracted ray. The equation for this is tan(thetax)= n2/n1 so for n=1, than tan(thetax) = 1.52 thetax= arcTan (1.52) thetax= 56.7 deg
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09:03:22 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 do understand how to do this problem.
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09:07:11 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 --> we know that we are dealing with 650 nm light, and since there are 28 lines, then m= 27 so the thickness is 27 * 1/2* 670 nm= 9000 nm
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09:07:49 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 do understand how to solve for the thickness
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09:14:06 **** gen phy how many wavelengths comprise the thickness of the foil?
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RESPONSE --> I think you would use the following equation to solve for how many wavelengths comprise the thickness of the foil: you first have to solve fo the thickness, which would be: t=1/2 * 2*650 nm =9.05*10^3nm= so to find how many wavelengths you would use this equation: 2t=mlambda 2(9.05 * 10^3 nm)
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09:15:14 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 understand that the 2(9.05 um)=m(6.70 * 10^7m) and that you convert all units to meters m=27 wavelengths
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