course PHY »[ã\›ã¼¤ú¯Ì†µ‘¤Í˜¸h³—‡ŠÂassignment #030
......!!!!!!!!...................................
19:57:23 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)?
......!!!!!!!!...................................
RESPONSE --> Tangent of the polarzing angle=n. In this case n=1.52. So by solving for the arc tangent of 1.52, so the degree is 56.65.
.................................................
......!!!!!!!!...................................
19:57:31 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.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:45:46 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?
......!!!!!!!!...................................
RESPONSE --> The Thickness of the gap would be determined by 2(thickness)=(wavelength), where wavelength is 650/1.52=427. So t=427/2= 213 meters.
.................................................
......!!!!!!!!...................................
20:46:59 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) **
......!!!!!!!!...................................
RESPONSE --> Oops, ok I did not need to divide my wavelength by the index refraction of glass first. I also left out my m. Understand it from there...
.................................................
......!!!!!!!!...................................
20:51:08 **** gen phy how many wavelengths comprise the thickness of the foil?
......!!!!!!!!...................................
RESPONSE --> I think my equation before can be used to solve for wavelengths and I would solve for m. So using the answer given in the above problem, I would have the equation 2(t)=m(wavelength) and solve for m. so 2(9000 nm)=m(670) So m=26.86
.................................................
......!!!!!!!!...................................
20:51:17 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.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
"