Intensity for Young’s Double Slit Interference

[Aprtaenl]

Homework Statement

Hi

The experimental setup involved a helium neon laser of wavelength 670nm +- 1nm, pointed towards a lens with focal length 500mm +- 1mm. (so that the equations used corresponded to the far field diffraction pattern) and onto a photo diode so that for all intents and purposes the voltage is proportional to the Intensity.

For a diffraction coarse grating consisting of a large number of slits, i collected data approximately 30 point.

Homework Equations

I am using this equation
Intensity Formula.png

which is a product of double slit interference and single slit diffraction.

From this I what to determine the finite distance of the slits (a) and the distance between the slit separation (d).

I believe I also need to use the equations which describe the distance between successive maxima and minima respectively.

maxima and minima equations.png

Can someone please explain how to obtain the values of a and d, from these equations and (potentially my data plot, additionally if anyone has any knowledge on python and could tell me what kind of function i need (i think it might be sin^2x multiplied by a gaussian envelope) but am not sure.

The Attempt at a Solution

[/B]
I am attempting to plot a graph of the intensity as a function of distance on python but have not been able to produce the fit yet, so am attaching a dot to dot fit produced on excel to contextualise the results, where on the x-axis i have position and on the y-axis i have intensity(voltage).

 

[BvU]

Hello Aprtaenl,

I hope you understand your account is far from complete: laser, lens, photodiode. Where ? And then, as a kind of afterthought, a grating as well. Where ?

Then some cut/paste equations with variables that do not appear anywhere else ("all variables and given/known data"?)

A chart with no axis titles (1.8 candelas? 10 meter ?)

Bit hard to assist under such conditions, I'm afraid ! Please have pity on us poor helpers !

Perhaps reading the guidelines is a good idea for you ?

 

[Aprtaenl]

Thank you for responding BvU
I see your point about making things clearer.

Experimental Set up in 2d.png

To start with i have attached a rough 2d sketchup of the experimental setup mounted onto an optical bench.

The equation I have used

Intensity Formula.png

Is based on the intensity distribution for single slit diffraction, with the assumptions that there the slit width d > > lambda (wavelength of light) and that f > > d (quantities defined below).

And the formula for double slit interference

which i think arises from malus' law.

I(x) is intensity as a function of position
I₀ is a the inital intensity (5v)
a is the width of the slit
d is the separation between two adjacent slits
lambda is the wavelength of the monchromatic helium neon laser, of wavelength 670nm
f is the distance between the lens and the photodiode - 500mm
x is the perpendicular distance between the normal and the fringe

The equation for the Intensity formula, is a product of the previous two equations. and should map to this graph if data is collected and plotted.

The Graph that I have drawn based on my data has Voltage (Intensity) in volts on the Y axis and position on the X axis in mm

1) My first question is how would I convert my distances into Radians, my guess is I would have to use sin theta =x/f, but i am unsure how to convert each data value as, x refers to the perpendicular distance between the normal and the fringe?

2) Using the equations, how would I go about determining the width of the slit (a) and distance between the slits (d)? Would it be beneficial to plot the graph as shown above?
Or would it be not that useful, as i don't think the gradient or area represent any meaningful quantities, and that it is simply used for the purposes of a visual representation

3) if possible would it be possible to help me fit my data points to a curve in python. I have tried fitting a sine squared graph, but the fit does not take into account the different sizes of the maxima, I have been experimenting with a Gaussian envelope multplied by the sin squared function. And having been trying to make the amplitude dependent on another function but to no avail.

Any suggestions would be greatly appreciated

 

[ehild]

Homework Statement

[/B]
For a diffraction coarse grating consisting of a large number of slits, i collected data approximately 30 point.

Homework Equations

I am using this equation
Intensity Formula.png
which is a product of double slit interference and single slit diffraction.

You said you measured with a multi--slit diffraction grating, but used the formula for Young double-slit experiment. It is not appropriate for your experiment. See
https://sites.uAlberta.ca/~pogosyan/teaching/PHYS_130/FALL_2010/lectures/lect36/lecture36.html

 

[BvU]

Ersbeth is awake early !

Aprtaenl (how do you pronounce that?): My compliments for the improvements between posts #1 and 3; much clearer.

The equation for the Intensity formula, is a product of the previous two equations. and should map to this graph if data is collected and plotted.

As Ehild says: your laser beam illuminates more than two slits (as witnessed by the intensity profile). Check e.g. hyperphysics for the differences going from 1 to 2, 3 etc. slits.

my guess is I would have to use sin theta =x/f

In physics we try to avoid the term 'guess', so make a drawing and 'conclude'

• $\tan \theta =x/f$ (at least if you focused accurately and the focal length is indeed given corretly -- but of course you also did a careful measurement of the distance, I expect ?
• $\tan \theta \approx\theta$ to a very good approximation

And then realize that $\theta$ doesn't even appear in your equations
You also want to worry if f is the distance from grating to photodiode or the distance from lens to photodiode
And where the diode sensitive area is located exactly in the axial direction
And how wide the entrance slit of the photodiode is (this limits your resolution in $\theta$)

My advice: think physics first and foremost. Do pen and paper analysis straightaway to discover weak points in the analysis and to find preliminary values for the features you are after and sort out their relevance.
In this case the number of lines/mm is preeminent. Your maxima are 1.2 mm apart so you can immediately obtain a first estimate of that.
Then with the beam diameter that you measured you estimate the number of illuminated slits to see if the single slit envelope is relevant or not. (With only three peaks getting the slit width is harder to establish accurately anyway)

Python and other fancy stuff can be done later and by anyone.

 

[Aprtaenl]

You said you measured with a multi--slit diffraction grating, but used the formula for Young double-slit experiment. It is not appropriate for your experiment. See
https://sites.uAlberta.ca/~pogosyan/teaching/PHYS_130/FALL_2010/lectures/lect36/lecture36.html

Thank you for the response ehild

After reading through the page I saw that the formula for N slits was

As oppose to my formula for two slits.

The website denotes p as the pairs of slits.

I am ensure to what I would use for this value.

I assume N is the total number of slits.

I assume that I can use the same angle assumption and replace theta with x/f (as it does not reference this on the website).

However, even with the updated formula I am still confused as to how to work out the values of a and d.
Would modelling the graph help?

 

[BvU]

Would modelling the graph help?

No. That's for later -- details, details.

 

[Aprtaenl]

No. That's for later -- details, details.

820e81fa-82ad-42c7-83c8-308a7a93ec77.jpg
dbd232a6-a131-47d5-8972-fea5c6186eb2.jpg
I have calculated an approximate value for d, and asked my lab coordinator for the actual value of the coarse grating, which was 4000 lines per m, so my side comment is irrelevant. I found the manufacturer's beam width for the laser which was 1mm. and so approx. 3/4 slits would be illuminated at anyone time meaning that the single slit envelope should still be relevant

 

[BvU]

Congratulations with your first like !
You did not let your expectation (100 lines/mm) interfere with your experimental result, which is commendable.

I do have some difficulty to understand your error analysis:

• Even when roughly looking at the graph of the interference pattern, I don't see an error of 0.3 mm coming up.
• The wavelength is 670 nm (not 1/670 !) and apparently you were given a standard deviation of 1 nm; that is only 0.14 %, not 14 %.

(I seem to remember a HeNe laser has 632.8 nm, but perhaps you have been given a laser diode ?)​

So the 1.2 mm is the main source of inaccuracy in $d$ -- and that you can perhaps improve a bit using a fit on the profile.

I also expect the beam diameter to be around 1 mm -- but this is deceptive: our eyes respond logarithmically so we tend to overestimate. Measurement would be better ! The profile is gaussian and circular symmetric (meaning not all of the beam illuminates $\approx$ 3.6 lines.

The expected diffraction pattern on the basis of this primary result would therefore be something like what you see here and it will take some explaining to match that with the pattern you observed ... in particular the two similar peaks instead of the expected central peak

Did you visually observe that too ? And: why did you only record three peaks ? As you can see from the single slit envelope, there will generally be a few peaks missing, yielding information about the slit witdh

What about my remark about the position of the grating on the axis ? And the photodiode (was it equipped with a slit, was it really very small, etc.) ?

 

[Aprtaenl]

Congratulations with your first like !
You did not let your expectation (100 lines/mm) interfere with your experimental result, which is commendable.

I do have some difficulty to understand your error analysis:

• Even when roughly looking at the graph of the interference pattern, I don't see an error of 0.3 mm coming up.
• The wavelength is 670 nm (not 1/670 !) and apparently you were given a standard deviation of 1 nm; that is only 0.14 %, not 14 %.

(I seem to remember a HeNe laser has 632.8 nm, but perhaps you have been given a laser diode ?)​

So the 1.2 mm is the main source of inaccuracy in $d$ -- and that you can perhaps improve a bit using a fit on the profile.

I also expect the beam diameter to be around 1 mm -- but this is deceptive: our eyes respond logarithmically so we tend to overestimate. Measurement would be better ! The profile is gaussian and circular symmetric (meaning not all of the beam illuminates $\approx$ 3.6 lines.

The expected diffraction pattern on the basis of this primary result would therefore be something like what you see here and it will take some explaining to match that with the pattern you observed ... in particular the two similar peaks instead of the expected central peak

Did you visually observe that too ? And: why did you only record three peaks ? As you can see from the single slit envelope, there will generally be a few peaks missing, yielding information about the slit witdh

What about my remark about the position of the grating on the axis ? And the photodiode (was it equipped with a slit, was it really very small, etc.) ?

Error analysis
sorry for typing 0.3mm, i was looking at the wrong points, the distance between maximum 1 and 2 is 1.25mm and between maxima 2 and 3 it is 1.2mm, so an error of 0.05mm. thank you for clearing up the error on the wavelength.

After checking with the lab manual i see that we used a laser dioide not HeNe which would explain the wavelength.
When making visual observations of the pattern on the screen, it was difficult to tell at which point the central maxima occurred, and whether it is much brighter than the surrounding maxima. Even though I understand the theory states that the central maxima should have the highest intensity.

I was only able to take three peaks, due to time constraints in the lab, as we had to perform this experiment for 2 more diffraction gratings, and so our lab demonstrator said taking 3 peaks and troughs with data points in between, would most likely be the best course of action. (I have attached images of the other 2 diffraction gratings)

The position of the grating on the axis was aligned with with the laser and was positioned between the laser and the lens. So that when the laser passed from the grating into the center of the lens. The height of the laser was measured on the optical bench and the diffraction grating's mount was adjusted until it was a similar height, verified with a ruler. This was done to ensure that the laser was focused into the small aperture on the photodiode. The photodiode was on a transition stage, and the position of it was adjusted with micrometer to ensure that it aligned with a maximum, this was verified with the voltmeter, to ensure that it was actually a maximum by checking points either side of the maxima to ensure they started to decrease.

The size of the aperture on the photodiode was circular, and very small, i did not measure the aperture opening but from my recollection looked to be 2-3mm wide EDIT [ I found in the appendix of my lab manual the slit to be 100 micrometers]