# Homework Help: Single Slit / Wavelength / Central Maximum

1. Mar 26, 2015

### julianwitkowski

1. The problem statement, all variables and given/known data

A laser emitting light with a wavelength of 560 nm is directed at a single slit, producing an interference pattern on a screen that is 3.0 m away. The central maximum is 5.0 cm wide. Determine the width of the slit.

2. Relevant equations

d = λ / sin(Θ)
tan(Θ) = x / L

or...

w = λL / y

3. The attempt at a solution

Attempt #1

d = slit width
λ = wavelength (560nm = 5.6· 10⁻⁷m)
Θ = angle

If the central maximum is 5.0cm wide, then is the distance from the center to the first minimum always half of that or 2.5cm?

tan(Θ) = 0.025m / 3m
Θ = tan⁻¹ (0.025m / 3m) = 0.477°
d · sin(Θ) = λ

d = λ / sin(Θ) = 560nm /sin(0.477°) = 5.6· 10⁻⁷m / sin(0.477°) = 6.72 · 10⁻⁵ m = 67200 nm

Attempt #2

d = slit width
λ = wavelength (560nm = 5.6· 10⁻⁷m)
L = distance to screen.
y = central maximum.

w = λL / y = 5.6 ·10⁻⁷m · 3m / 0.05m = 3.36 · 10⁻⁵ m = 33600nm

2. Mar 26, 2015

### rude man

Yes. the pattern is symmetrical about θ. sinθ is an odd function. So is tanθ.
Stop right there!
Not sure what your y is. Is it the total width of the central maximum? If so the formula is incorrect.
Your 1st formula derivation is not trivial so you should just accept it unless you're afraid you'll have to derive it in a test someday.

3. Mar 26, 2015

### julianwitkowski

I wasn't sure if I was supposed to cut the central maximum in half or not.
Thank you for your time, this has helped me :)

4. Mar 26, 2015

### BvU

Hello Julian,

As Rudy indicates, the formal derivation of the single slit diffraction pattern isn't trivial. And personally I don't like the little arrows addition approach (matter of taste, I suppose). Better to study that using Fourier transformation.
Since I think you are a curious person (and also because you ask for understanding in post #1), this is how they taught me before I knew about Fourier transforms:

According to the Huygens principle, all points in the slit opening function as point sources for the emanating wave.

If you divide the opening in two halves , then for rays that go in the direction of the first minimum from the top half (#3 is shown in the drawing), there is a corresponding ray in the lower half with which it interferes destructively (#4 for #3 and then stepping down until the top one is in the center and the corresponding lower one at the lower end of the slit), i.e. a ray that has a phase difference of $\lambda/2$. So you get $d/2 \; \sin\theta = \lambda/2\;$ for the first minimum.

You can do the same thing for subdividing in four quarters to get $d/4 \; \sin\theta = \lambda/2 \;$ for the second minimum.

In between is a (relative) maximum, because if you divide in three parts, two of those will interfere destructively and that leaves the contribution of the remaining third