How many interference fringes are visible in the central diffraction peak if the ratio of slit separation to slit width is 3.125?
d*sin(θ) = m*λ for interference maxima with slit separation d, maxima are at m = 0, 1, 2, 3 ...
a*sin(θ) = m*λ for diffraction minima with slit width a, minima are at m = 1, 2, 3 ...
The Attempt at a Solution
md = m for diffraction minima
ma = m for interference maxima
ma/md = 3.125
d*sin(θ) = md*λ
a*sin(θ) = ma*λ
-> d/a = (md/ma)*λ
When md = 1, ma = 3.125, so the central diffraction peak contains the 3rd order interference maxima. This means that there are 3 interference maxima on the right of the central interference fringe, and 3 on the left, giving a total of 7.
My instructor gave a formula to calculate the number of interference fringes visible in the central diffraction peak: 2(d/a) - 1
Using this, I get 2(3.125) - 1 = 5.25, or after rounding, 5 peaks. What am I missing?
Also, this formula allows for an even number of interference fringes in the central peak (consider d/a = 2.5, 2(2.5 - 1) = 4), is that possible?