Table of Contents
Diffraction - At Home Makeup Lab
This exercise will introduce you to the phenomena of single slit interference.
The exercise is designed to be done remotely, there is no in-lab component.
Read through the theory section about interference and diffraction. At the end of this section we describe an apparatus which was used to scan the intensity of light which has passed through a single slit. You are then given a set of pre-recorded data from which you will make measurements to the best of your ability. Carefully consider the uncertainty in your measured values and describe in detail how you performed the measurements.
You will be graded on the lab notebook which you will submit to your TA in the usual way.
Lab report template
The link below will ask you to sign-in to your UChicago Google Drive and will create a copy of the file for you to edit.
Theory
Diffraction is a consequence of the principles of superposition and interference. Click below for a short review of superposition.
Superposition
Interference is simply a consequence of the principle of superposition. This means that at any point in space, the total amplitude of the disturbance due to two separate traveling waves is the sum, or superposition, of the amplitudes of the two separate waves.
Mathematically, we can write two sinusoidal traveling waves of equal amplitude, $y_1$ and $y_2$, at point $x = 0$ as
| $y_1 = A\cos (\omega t + \varphi_1)$ $y_2 = A\cos (\omega t + \varphi_2)$, | (1) |
where $\varphi_i$ may be a function of position, but not of time. The superposition of these waves yields the total wave $y_T$,
| $y_T = y_1 + y_2 = 2A\cos\left(\dfrac{\varphi_2-\varphi_1}{2}\right)\cos \left(\omega t + \dfrac{\varphi_2 + \varphi_1}{2}\right)$. | (2) |
This gives a total intensity of
| $I_t = \left< y_T^2\right> = 2A^2\cos^2\left(\dfrac{\varphi_2-\varphi_1}{2}\right) = 2A^2\cos^2\left(\dfrac{\Delta\varphi}{2}\right)$. | (3) |
Note that the total intensity depends critically on the phase difference $\Delta\varphi$ between the two waves. If $\Delta\varphi$ is zero or an integral multiple of $2\pi$, then the waves are in phase and $I_T$ is twice as large as the sum $I_1+I_2$ (constructive interference). If $\Delta\varphi =(2n+1)\pi$, where $n = 0, 1, 2, \dots$, then the waves are 180 degrees out of phase and $I_T=0$ (destructive interference).
A sinusoidal traveling wave, propagating in the $x$-direction and produced at $x_0$ with phase $\varphi_0$, has the form
| $Y = A \cos\left(\omega t - k(x-x_0)+\varphi_0 \right) = A \cos \left(2\pi\left(ft - \dfrac{x-x_0}{\lambda}\right) + \varphi_0\right)$, | (4) |
where
| $\omega = 2\pi f$ $k = \dfrac{2\pi}{\lambda}$ $v = f\lambda = \dfrac{\omega}{k}$ | (5) |
so that the phase at any point $x$ is $\varphi = \varphi_0 - k(x-x_0)$ If two such waves interfere, $\Delta\varphi$ at the point of interference will depend on the distance traveled by each wave.
For example, suppose two such waves with identical $\omega$ and $\varphi_0$ are produced on the $x$-axis at $x_1$ and $x_2$. Then, at a distant point $x$, the phase difference will be
| $\Delta \varphi = k(x_2-x_1) = 2\pi\left(\dfrac{x_2-x_1}{\lambda}\right)$ | (6) |
and the intensity of the total wave at $x$ will be
| $I_T = 2A^2\cos^2\left(\pi\left(\dfrac{x_2-x_1}{\lambda}\right)\right)$. | (7) |
This analysis applies equally to transverse waves (microwaves and light) and to longitudinal waves (sound). Note that the two waves must have identical frequency and a fixed phase difference at their point of production.
Interference from two point sources
Figure 1 represents the wave disturbance originating from two point sources, S1 and S2, separated by a distance $d$. By the superposition principle, the intensity observed at the detector will be a maximum (i.e., the waves will be in phase) if the path difference, $\Delta r = r_2-r_1$, is an integral number of wavelengths, i.e. $\Delta r = n\lambda$. But, for a detector far from the sources, $\Delta r = d\sin\theta$. Therefore, the angles at which the intensity will be a maximum, denoted by $\theta_{max}$, obey the following relation:
| $n\lambda = d\sin\theta_{max} (\mathrm{for\;n = 0,1,2,\dots)}$. | (1) |
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| Figure 1: Interference from two point sources |
The intensity observed at the detector will be a minimum if the two sources are out of phase, i.e., $r = \left(n + \frac{1}{2}\right)\lambda$. Therefore, denoting angles where we expect an intensity minimum by $\theta_{min}$, we have
| $\left(n+\dfrac{1}{2}\right)\lambda = d\sin\theta_{min} (\mathrm{for\;n = 0,1,2,\dots)}$. | (2) |
Single slit diffraction
Now, consider an extended source, e.g., a slit, having width $a$ as shown in Fig. 2. We can consider the pattern of diffracted light coming from such an extended source as that resulting from the interference pattern of an infinite number of point sources within the slit, all oscillating in phase.
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| Figure 2: Single slit diffraction |
To find the angles where the intensity will be a minimum, let us divide the extended source of width $a$ into two equal regions: the top and bottom halves of the slit. Now suppose that we are at such an angle $\theta$ that the path difference between a source at the top of the slit and one at the middle of the slit is exactly one-half wavelength; then it is clear that these two will cancel and give no contribution at the detector. Also, it should be clear that the contribution from the entire slit will as well be zero at this angle. This is because for any source in the top half of the slit, we can find one in the bottom half (namely one a distance $a/2$ from the upper source) which will, when added, give zero intensity. Thus, our condition for a minimum intensity becomes
| $\Delta r = \dfrac{a}{2}\sin\theta_{min} = \dfrac{n\lambda}{2}$ (for $n = 1,2,3,\dots$). | (3) |
(Note that $n=0$ is not included.) Thus,
| $a\sin\theta_{min} = n\lambda$ (for $n = 1,2,3,\dots$). | (4) |
The maxima will occur approximately halfway between the minima. A single slit diffraction pattern is shown in Fig. 3.
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| Figure 3: Single slit diffraction pattern |
Two slit diffraction
The pattern of transmission through two slits will depend upon both the width of the slits and their spacing. The amplitude from either slit at angle is governed by the single slit diffraction pattern - whereas the intensity from the two interfering slits will depend upon the interference pattern from two point sources. An intensity pattern similar to that shown in Fig. 4 will result. Note that the two slit pattern of Fig. 4 would lie entirely beneath the single slit pattern of Fig. 3, showing that the larger scale shape of the two slit pattern is still dictated by the widths of the individual slits.
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| Figure 4: Two-slit diffraction pattern (with slit separation equal to three times the slit width) |
Apparatus and data
To create diffraction patterns, we directed a laser through a specially-designed slit holder and onto a white screen. (See Fig. 8) In front of the screen, we mounted a photodiode (light sensor) attached to a slide potentiometer (a variable resistor) so that we can measure the light intensity as a function of position along the screen. Both the intensity and detector position are read out and recorded on a computer. In addition, we have a gain knob which allows us to amplify the signal from the photodiode before sending it to the computer.
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| Figure 5: Laser with diffraction scanner |
We set up our photodetector a distance $L = 71.0 \pm 0.2$ cm from the diffraction slit holder and project the output onto our photodetector. The diffraction pattern we get when we pass the light through a single slit of width $a = 0.08$ mm is shown in Figure 9. In Fig. 9(a), we turn the voltage gain down so that we can see (almost) the entire central maximum peak. In Fig. 9(b), we increase the voltage gain so that we can better see the smaller amplitude maxima on either side. (The flat top to the central maximum is because the detector can put out only a maximum of 5 V. We cannot amplify the signal beyond that point.)
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| (a) Single slit, zoomed-out | (b) Single slit, zoomed-in |
| Figure 6: Light intensity (arbitrary units) of the diffraction pattern produced by single slit. (Click to enlarge.) | |
Exercise
Determine the wavelength of the laser from the diffraction pattern.
From the data provided in Fig. 6 for the single slit diffraction, fill out the following table, in your notebook, for the first 4 minima on either side of the central maximum. In order to calculate the minimum angle, use the fact that $\theta_{\textrm{min}}$ is one of the angles in the right triangle with sides $L = 71.0 \pm 0.2$ cm and $x_{\textrm{min}}$.
A note on measurements and uncertainties.
You are working with pre-recorded data. There is nothing unusual about this in physics research, as there are many examples of experiments where you work with data that was collected by others. Large particle detectors, instruments aboard satellites, etc. are all examples of experiments where data are collected and shared within a collaboration of scientists.
For this makeup lab you are given plots of the data for two scans, one where the sensitivity of the detector is low enough to see the highest peak, and another with the gain turned up to improve the visibility of the weaker peaks. You will have to figure out your own method of estimating the locations of the features you need to measure, and you will have to think carefully about how well you are able to estimate these values. What is most important is that you decide upon a method which allows you to obtain repeatable results. It is acknowledged that you can only do so much with the data provided in this form, but you should strive to make the best measurements possible. Some suggestions include:
- Zoom in on the graphs and use the cursor to estimate values.
- You can print out the plots and use a pencil and ruler to make measurements.
- You could save the images and open them in an image processing program that allows you to read off pixel values.
- You have two separate scans, at two different gains. Some of the features can be measured on both plots and their values combined.
- You can get some estimate of the repeatability of your method by measuring the same feature
In your lab notebook you need to clearly describe how you made the measurements and how you estimated their uncertainties.
| diffraction order, $n$ | location of the left minimum, $x_{\textrm{min,left}}$ (cm) | location of the right minimum, $x_{\textrm{min,right}}$ (cm) | average distance from the center, $x_{\textrm{min}} = (x_{\textrm{min,right}} - x_{\textrm{min, left}})/2$ (cm) | $\sin\theta_{\textrm{min}}$ | $a\sin\theta_{\textrm{min}}$ (cm) |
|---|---|---|---|---|---|
| 1 | |||||
| 2 | |||||
| 3 | |||||
| 4 |
- Now, use this data to determine the laser wavelength, $\lambda$. You can use whatever method you like (and explain), but it should incorporate all four measured distances.
- How does your value compare to the literature value of $\lambda$ = 632.8 nm?