Physics question need hep here!! on waves!!?

Question

Consider a two slit interference pattern, for which the intensity distribution is given by

I = I(not) cos^2 (pi*d*sin(theta) / lamda)

Let θm be the angular position of the mth bright fringe, where the intensity is I(not). Assume that θm is small. Let θm+ and θm- be the two angles on either side of θm for which I=½I(not). The quantity Δθm= |θm+ -θm-| is the half-width of the mth fringe. Calculate Δθm. How does Δθm depend on m?

Answer 1

Let's represent θm+ as Δθ, I(not) as Io, the I value at θm+Δθ as IΔ, and the intensity relation by
Io = cos^2(k*sin(θm)) = 2IΔ = 2cos^2(k*sin(θm+Δθ)),
where k = pi*d/lambda
Then (I'm doing this step-by-step to (I hope) avoid errors)
Io/2 = cos^2(k*sin(θm+Δθ))
sqrt(Io/2) = cos(k*sin(θm+Δθ))
arccos(sqrt(Io/2)) = k*sin(θm+Δθ)
arccos(sqrt(Io/2))/k = sin(θm+Δθ)
arcsin(arccos(sqrt(Io/2))/k) = θm+Δθ
arcsin(arccos(sqrt(Io/2))/k)-θm = Δθ
Δθm = 2Δθ = 2(arcsin(arccos(sqrt(Io/2))/k)-θm)
I think Δθm is independent of m.
EDIT: I tested the Δθ equation, solving for Δθm and checking for IΔ = Io/2 and it works for multiple values of m; i.e., integer multiples of a theta value that yields a peak. Surprisingly (to me), Δθm is not independent of m.

Answer 2

you will hear beats and the beat frequency often is the version between the two frequencies case in point, if a 440Hz and 438Hz sound are emitted, you will hear beats with a frequency of 2Hz if the amplitudes are lots diverse you could desire to no longer hear the consequences of the 2nd tone too nicely

Answer 3

http://en.wikipedia.org/wiki/Double-slit_experiment