Calculating Wavelength of Laser Diffracted by Slit Patterns

[Oijl]

Homework Statement

I am to calculated the wavelength of the laser I shone through different slit patterns to create diffraction patters on a screen.


For the single slit diffraction,
Data:
xm is the distance from the central image to the mth minimum.
x1 0.008
x2 0.015
x3 0.023
x4 0.030
x5 0.037
The distance L from the grating to the screen is 1.865
The slit width a is 0.00016

Homework Equations

asinø=my
Equation for minima for a single slit system

The Attempt at a Solution

I graphed sinø verses m; in other words:
sinø = (y/a)m
So the slope of that graph is y/a.
So the slope times a is y.

[[ I calculated sinø using (xm)/( ((xm^2)+(L^2))^(1/2) ) ]]

The slope, I found using Excel, of the graph of sinø vs m, is 0.003.
So lambda, y, should be
y = 0.003*0.00016 = 4.8*10^-7
This is a wavelength of 480nm.
The literature value is close to 640nm.

My results for the double-slit is even more off, giving me a calculated wavelength of 2000nm - which is obviously wrong because I wouldn't have been able to see that.

What am I doing wrong?

 

[rl.bhat]

Your calculation of the is wrong.
The slope of the graph is the band width β.
From the observation what is (x3 - x1) ? What is (x4 - x2) ?
From these values find β.
Here values of x are small compared to L. So sinθ is nearly equal to θ = x/L.
Now solve for λ.

 

[Oijl]

Okay, I don't know what you mean. Band width B?

What is B for a single-slit system?

For a double-slit system?

For a grating?

 

[rl.bhat]

Distance between the successive minimum or maximum is called band width.
From the graph, the slope is the band width. According to me it is 0.0075 m.

 

[Oijl]

I'm sorry, but I still don't understand.

The y-values on my graph are sin(theta) and the x-values are m. The slope of this is, you say, B? And B is the distance between successive minima?

Then how does the slope of sin(theta) vs m relate to wavelength? How does it relate to slit width? How would this be different for a double-slit, meaning, how does the slope of sin(theta) vs. m for a double-slit diffraction pattern relate to the slit width and slit separation?

 

[Oijl]

Maybe, to help you understand where I am, here are the equations I was given with which to do this:

I = Io(sinB/B)^2
B = ((πa)/y)sinø
For minima, sinB = 0, so B = mπ
And so
mπ = ((πa)/y)sinø for minima
which is the same as
asinø=my

 

[Oijl]

The last post is the equations for the single-slit. Here are the equations I was given for the double-slit:

I = 4Io((sinß/ß)^2)cos^2(alpha)
alpha = ((πd)/y)sinø
When cos(alpha) = 0 there are minima, and
alpha = (m + 1/2)π = ((πd)/y)sinø
which is, rearranged,
dsinø = (m+1/2)y

 

[rl.bhat]

Will you show me the calculations of sinφ.
According to me, since φ is very small, sinφ is nearly equal to φ, which is equal to xm/L.
Αnd from your observations slope should be 0.00402. Check it.

 

[Oijl]

I calculated sinø using (xm)/( ((xm^2)+(L^2))^(1/2) )

So my curve is (xm)/( ((xm^2)+(L^2))^(1/2) ) = (slope)m

I'm told to use this curve to find, for the single-slit, the wavelength and slit width, and, for the double-slit, the slit width and separation.

Therefore, I should be able to express the slope in a relationship with lambda, slit width a, and slit separation d.

I can't reason out what that equation, slope = [y][a][d], is.

I also had used the approximation sinø = ø = xm/L, but that didn't change the slope within three significant figures.

I'm calculating the slope using a best-fit line in Excel, and it gets 0.003. I've checked my formula I put in the cells, and it is the correct formula for sinø.And, by the way, thank you for sticking with me through this. It's due in an hour, and I'm frustrated at my mental thickness about this.

 

[rl.bhat]

I hope you have got the straight line.
Slope of a straight line can be written as (y2 - y1)/( x2 - x1)
In your graph it can write down as
(x3/L - x1/L)/(3 - 1) or (x4/L - x2/L)/(4 - 1) which is constant.
Equate it to λ/a and find λ.
Your double slit formula is correct.

 

[Oijl]

Ah. The numerical value for my slope was incorrect (as you said).
But at least I was right that the slope is λ/a.
Well, thanks!