# Homework Help: Distance between maxima of two-slit interference

1. May 21, 2013

### PhizKid

1. The problem statement, all variables and given/known data
A light of wavelength 500 nm interferes through a two slit experiment on a screen. The distance between the first minimum and fourth minimum is 1.68 cm. Find the distance between the central maximum and first order maximum.

2. Relevant equations
d*sin(θ) = nλ = Δ

d = distance between slits
Δ = path difference

sin(θ) = tan(θ)
tan(θ) = x / L

x = distance between orders
L = distance between slits and screen

3. The attempt at a solution

We can find the path difference between the first and fourth minimum by:

(3.5)λ which is 0.000175 cm. (I converted 500 nm to 0.00005 cm).

So we can get: $\frac{0.000175 cm}{d} = \frac{1.68 cm}{L} = tan(θ_1) = sin(θ_1)$.

We need to find: $\frac{0.00005 cm}{d} = \frac{x}{L} = tan(θ_2) = sin(θ_2)$

But it doesn't look like we have enough information. How can I solve this? Can I use the ratio d / L to set the equation $\frac{0.000175}{1.68} = \frac{0.00005}{x}$?

2. May 21, 2013

### barryj

Things to think about: Does the wavelength of 500 nm really matter? is the distance between minimums the same as between maximums? Is the first order max next to the central max?"

3. May 21, 2013

### PhizKid

I found the distance between minimums is larger than the distancce between maximums. And no, the first order max is not next to the central max because they are separated by a minimum

4. May 21, 2013

### barryj

Of course there is a minimum between the central max and the first order max. If you look at the pattern, I think you would see a central bright line, then a dark line, then another bright line, then a dark line and etc, yes. As I recall, the distances get a little larger as you move further away from the center but I doubt this is of concern here.