- #1
jmm5872
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I am trying to calculate how wide of a screen you would need in order to observe an interference pattern produced by Young's original real-life double slit experiment. He cut a single narrow slit (width ds) in a window shade, admitting a narrow sliver of sunlight into a dark room. He inserted a colored filter into the sunbeam that transmitted a narrow band \Delta\lambda of wavelengths around a center wavelength \lambda0. A distance Ls away, the filtered beam illuminated an opaque screen in which he had cut 2 identical slits, each of width w, separated by distance d > w. He observed the interference pattern on a screen located distance L beyond the slits.
Ls = 2 m
w = 0.1mm
d = 0.25 mm
\lambda0 = 0.5 microns
L = 4 m
r = radius of pattern (or screen needed)
Here is what I did, and my results seem too small.
\theta = 1.22(\lambda/d) = r/L
Solving for r I got:
r = 1.22(L\lambda/d) = 4.88 mm
Like I said, this doesn't seem correct. It seems like the entire pattern will spread out more. How can I determine the width of the pattern?
Ls = 2 m
w = 0.1mm
d = 0.25 mm
\lambda0 = 0.5 microns
L = 4 m
r = radius of pattern (or screen needed)
Here is what I did, and my results seem too small.
\theta = 1.22(\lambda/d) = r/L
Solving for r I got:
r = 1.22(L\lambda/d) = 4.88 mm
Like I said, this doesn't seem correct. It seems like the entire pattern will spread out more. How can I determine the width of the pattern?