#$&* course Phy 202 Sunday, July 22, 2012, between 11:20 PM and 11:25 PMI'm sorry that I had such a hard time with the second problem; the book didn't give very explicit instructions re: how to count the number of intervals from the number of bands (in hindsight: If both the high end and the low end are light, count intervals by counting bands but starting with zero, i.e., Number:Intervals = Number:Bands - 1.
.............................................
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 theta_p = 56.7 degrees, approximately. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - SOLUTION: ???? Does ""28 lines"" mean that the plate's appearance is a) white at both the highest and the lowest points with 28 ""lines"" visible, for a total of 28 light-to-dark cycles + 1 half-cycle = 28.5 cycles; b1) white at the highest or lowest point and dark at the other, with the exterior dark portion counting as a ""line,"" for 28 light-to-dark or dark-to-light cycles; b2) white at the high or low end and dark at the other, but with the outer dark portion *not* counting as a ""line,"" for a total of 29 dark-to-light cycles (i.e., 28 ""lines"" plus the dark part at one side); c1) dark at both the high and the low end, with each counting as a ""line,"" leaving 27 dark-to-light cycles plus one ""half-cycle"" for a total of 27.5 cycles; c2) dark at both the high and the low end, but with one and only one of the top or bottom counting as a ""line,"" for 27 cycles even [Hindsight: This would also be the number arrived at if it were light on both sides and interval numbering starts by counting the first band at 0; see below]; or c3) dark at both the high and the low end, but with neither counting as a ""line,"" for 28 dark-to-light cycles plus one ""half-cycle"" for a total of 28.5 cycles? ???? - - In any event, formula to use is 2 * thickness = Number:Cycles * wavelength; therefore, thickness = (Number:Cycles / 2) * wavelength. - - a) Thickness = (28.5 cycles / 2) * 670 nm per cycle = 9547.5 nm or 9.5475 micrometers. - - b1) Thickness = (28 cycles / 2) * 670 nm per cycle = 9380 nm or 9.380 micrometers. - - b2) Thickness = (29 cycles / 2) * 670 nm per cycle = 9715 nm or 9.715 micrometers. - - c1) Thickness = (27.5 cycles / 2) * 670 nm per cycle = 9212.5 nm or 9.2125 micrometers. - - c2) Thickness = (27 cycles / 2) * 670 nm per cycle = 9045 nm or 9.045 micrometers. - - c3) Thickness = (28.5 cycles / 2) * 670 nm per cycle = 9547.5 nm or 9.5475 micrometers. confidence rating #$&*:e: computations; 1.5 to 2 re: interpretation of language in problem (re: what qualifies as a ""line"")
.............................................
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. ???? OK, so interval numbering a) counts each ""down and back"" as one interval and b) starts with the first dark line corresponding to zero, meaning that c2) above is the most correct / least incorrect interpretation. I now get that much, but does a dark area at the top/""high"" side count as a line, and does a dark area at the bottom/""low"" side count as one? That is, I can solve a problem that tells me how many lines there are, but in practice (e.g., from looking at a photograph), I don't know what counts as a line and what doesn't. Also, neither this problem [in hindsight -- yet] nor the text says expressly whether a full wave cycle or a half wave cycle generates a dark-to-light-to-dark transition. ???? 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) ** ???? OK, that answers the second part; now all I need to know is whether dark portions at the high and/or low sides of the plate count as lines. ???? **** gen phy how many wavelengths comprise the thickness of the foil? - SOLUTION: If 2(thickness) = number of bands * wavelength = number of half-wavelengths * wavelength, then 2(thickness) = (2(number of wavelengths)) * wavelength, and thickness = number of wavelengths * wavelength. Therefore, number of wavelengths = thickness / wavelength, as one would expect when asked to express thickness in terms of number of wavelengths. - - Thickness / wavelength = number of waves ""covered"" by one foil thickness, i.e., how many wavelengths are equivalent to one ""foil-thickness."" - - - For a), 14.25 [wavelengths]; for b1), 14 even; for b2), 14.5; for c1), 13.75; for c2), 13.5; c3), 14.25. - - - If c2) or its ""in-hindsight"" numerical equivalent corresponds to the answer that you're looking for, as I infer from the above, then it's 13.5 wavelengths per thickness. 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. ???? But isn't m, the number of bands, the number of half-wavelengths, for the reason that each dark band is caused by a half-wavelength of separation? And wouldn't the number of full wavelengths be half the number of half-wavelengths and therefore m/2 rather than m, namely 27 / 2 = 13.5 rather than 27? ???? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 1.75 to 2.25 (depending on whether my alternative method for the second part is correct; in any event, I'm not pleased with my having to do a ""shotgun"" approach up front) ------------------------------------------------ Self-critique Rating: OK or 2 (depending on whether my understanding of the second part is correct)" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!