|Jan25-11, 04:14 PM||#1|
Light intensity through 2 slits
1. The problem statement, all variables and given/known data
Monochromatic light of wavelength 620nm is going through a very narrow gap through the first curtain,then encounters a second curtain that is parallel with the first in which there are two parallel narrow slits S1 and S2 as shown in Fig. Slit S1 is located at the point of the second curtain that is closest point S, and S2 is d away from the S1. At point P, which is equidistant from the S1 and S2 we measure the intensity of the light and get the same intensity in both cases when only one of the leaked S1 and S2 is open, while in the case when both are open, we get 3 times greater the intensity. Determine the minimum distance d of slits, when the slits S and S1 are at distance L = 12dm.
2. Relevant equations
path/vave length=phase difference/2*pi
The intensity is proportional to the electric wave squared
3. The attempt at a solution
I tried to solve this using phasors, but in the end I am left with the equation containing the unknown term of phase difference (which contains the distance between the slits) and the intensity of a single slit. Am I doing it right?
|Jan25-11, 06:11 PM||#2|
Welcome to PF! You have a difficult problem here and I am only offering a suggestion because it is about to fall off the first page and the real experts may not see it on page 2. If there was no phase difference for the two slits, you would get double the E field strength so, 4 times the intensity, wouldn't you? I'm sure you remember this better than I do! If so, then the next question is what phase difference will give 3 times the intensity or sqrt(3) times the E. It will be something like
E+E*sin(θ) = E*sqrt(3) I would think. Solve that for θ and then figure out what distance d will give you that phase angle for the S to S2 minus the S to S1 distance.
|intensity, light, phasors, slits|
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