- #1

Cade

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## Homework Statement

How many interference fringes are visible in the central diffraction peak if the ratio of slit separation to slit width is 3.125?

## Homework Equations

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?