course Phy 202 I downloaded the latest updated queries and I can't get asst 20 to work. It says subscript is out of range. I am using the wrong query program?I apologize for the delay in my assignments. I have a million excuses like every student, but I promise that it is not because I am being lazy. This weekend I have cancelled plans for Friday night and Saturday day so I can get caught up. I hope to finish this week's assignments 21-23 during that time. I bought my generator and will complete those labs this weekend. The cylindrical lens is giving me trouble in regards to the calculating the refraction. I have most of the lab finished and will submit what I have finished as soon as I can.
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15:49:34 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: arc tan of 1.52 = 56.65 deg
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15:51:06 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 --> tan (theta_p) = n2/n1 n2 is index of refraction n1 = 1 so tan (theta p) = n2/ 1= n2 tan (theta p) = 1.52 theta p = arctan (1.52) arctan 1.52 = 56.7 deg I did the work but didnt show it all.
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15:54:07 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 --> 28 lines are obsesrved normal 650 nm light the thickness would then be 28 lines * 1/2 (wavelengths) * 650 nm = 9100
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15:56:25 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 --> 28 dark lines, that means 27 intvervals. 1/2 * 27 * 650 nm = 8775 nm Well my calculations were close.
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15:57:28 **** gen phy how many wavelengths comprise the thickness of the foil?
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RESPONSE --> 2 * t = m' lambda 2 * (t) = m * 650 nm Where do I find t?
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15:58:53 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 --> Oh I use the 8775 nm for the thickness! How could I miss that? 2 * (8.775 um) = m (6.7 * 10^-7 m) m = 26.19 wavelengths. Okay.
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