Double Slit Interference Pattern

[jmm5872]

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 d_s) 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 $$\lambda$$₀. A distance L_s 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.

L_s = 2 m
w = 0.1mm
d = 0.25 mm
$$\lambda$$₀ = 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?

 

[kuruman]

$$\theta$$ = 1.22($$\lambda$$/d) = r/L

Where did you get this and why is it applicable here? For a treatment of the double slit and the correct equations, go to

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/slits.html#c1