Single Slit / Wavelength / Central Maximum

[julianwitkowski]

Homework Statement

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.

I have two ideas of the answer but I'm not sure which is right... Please help me understand which one and why.

Homework Equations

[/B]
d = λ / sin(Θ)
tan(Θ) = x / L

or...

w = λL / y

The Attempt at a Solution

[/B]
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

 

[rude man]

Homework Statement

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.

I have two ideas of the answer but I'm not sure which is right... Please help me understand which one and why.

Homework Equations

[/B]
d = λ / sin(Θ)
tan(Θ) = x / L

or...

w = λL / y

The Attempt at a Solution

[/B]
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?

Yes. the pattern is symmetrical about θ. sinθ is an odd function. So is tanθ.

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

Stop right there!

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

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.

 

[julianwitkowski]

Stop right there!
.

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 :)

 

[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