In this lab we will look at two closely-related phenomena: **wave interference** and **diffraction**. These topics have strong connections to the previous experiment on *standing waves* (which are the result of *interference* of forward and backward moving waves).

Unlike previous labs this year where you worked over two sessions towards one large goal, this lab is made up of several smaller investigations on different aspects of interference (in Part 1) and diffraction (in Part 2).

For each project, we will provide a lab template for you to use as your lab notebook. This file includes prompts that you will expand on as you go through the lab. 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.

To start, we will use the **Wave Interference** PhET simulation: https://phet.colorado.edu/sims/html/wave-interference/latest/wave-interference_en.html

When you open the lab, you have the option to choose a module.

- The
**Waves**module is a very basic simulation that creates waves from a single source. - The
**Interference**module looks at waves created by one or two coherent sources. - The
**Slits**module looks at waves created by a single plane-wave source as it encounters a barrier with one or two slits. (*We will not be using this module in this week's lab*.) - The
**Diffraction**module looks at light passing through differently-shaped apertures. (*We will not be using this module in this week's lab*.)

After selecting a module, the 4 options will now appear along the bottom of the screen so you can flip from one to another. (The previous module will be left undisturbed when you switch from one to the other.) There is also a **home** icon which returns you to the original selection screen.

Once again, the orange circle arrow in the lower right corner can be used to return the simulation to its original starting state.

We don't have to go over the mathematics of interference before we try to observe it, but we'll include a little bit here if you want to peek. Maybe try the simulation below and come back here if you need help understanding what's going on?

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.

To get a feeling for how the simulation works, go into **Waves** mode and create a continuous *light* wave.

Answer the following questions for yourself. *(No need to record them in your report. This is just practice.)*

- How can you measure the wavelength?
- Do you use the tape measure? Do you use the “two-channel scope”?

- How can you estimate the uncertainty?
- What are the uncertainties related to
*direct*measurements? - What are the uncertainties that get propagated through to calculated values?

- How can you lower the uncertainty on measured quantities?
- Can you measure one thing multiple ways? (For example, getting wavelength from peak-to-peak and trough-to-trough on the same wave?)
- Can you repeat measurements?
- Can you measure multiple cycles of a wave and divide by the number of cycles?

- Which color has the largest wavelength? Which has the smallest?
*Roughly*, what is the wavelength range (1 m to 10 m? 1 nm to 100 nm? Something else?)

Now switch to **Interference** mode and again select *light*. If you turn both sources on, you will see that the waves which they produce are **coherent**. That is to say, the two waves are emitted **in-phase** with each other; when the top source is, for example, at a maximum, so is the bottom source.

Now click the check boxes to turn on the *screen* and *intensity*.

When answering the following questions (and questions in future sections), you do not need to address each bullet point individually. It may be easier – and more concise! – to write a single answer after each group of questions that touches on all the things asked. Keep in mind that the notebook is your record of what you did in the lab and the questions are meant to guide you through your measurements and observations.

You probably want to include some screenshots in your notebook to help illustrate how and where you are making measurements. “Showing” is often more concise than writing out in words.

Choose a particular wavelength of light and a specific source separation (and record them in your notebook).

- Qualitatively (that is,
*without using numbers*) describe what you see on the screen. - Do you see maxima and minima? Do the positions of these maxima and minima move as you change the wavelength?
- If so, how? (Be specific. As you increase wavelength, what happens? As you decrease?)
- Do any of the maxima or minima stay put as you change wavelength?

- What happens as you increase or decrease the source separation? (Again, be specific.)

Now let's get quantitative. For one wavelength and separation, find a **maximum** (other than the central maximum) on the screen…

- Measure the light's wavelength and estimate the uncertainty.
- Measure the distance from each light source to the maximum and estimate the uncertainty in this distance.
- What is the difference between the distances from each source?
- How does this path length difference compare to the wavelength of the light?

Now, repeat for one **minimum** on the screen…

- Measure the distance from each light source to the minimum and estimate the uncertainty.
- What is the difference between the distances from each source?
- How does this path length difference compare to the wavelength of the light?

Putting all this investigation together…

What mathematical relationship between the two path lengths do we need to see constructive interference (a maximum) and what relationship do we need for destructive interference (a minimum)?

There have been a lot of student questions recently about why we ask you to estimate uncertainties *so much*. And again here, we've just asked you to estimate the uncertainty in your wavelength and in the measured distances between sources and maxima/minima. So why do we do this?

Whenever we compare values that have an inherent fuzziness – like the lengths that you measure here – the uncertainty helps us quantify how much difference is normal and how much should cause us to worry. At the simplest level, we can say that two numbers are “consistent with each other within uncertainties” (e.g. if you measure a value for the speed of light of $(3.05 \pm 0.10) x 10^8$ m/s and the literature value is $3.00 x 10^8$ m/s, you would say your measurement is “consistent with the literature value within uncertainties”), but there are more sophisticated methods we can use too (like the $t^{\prime}$ test we used in PHYS 131 when looking at pendulum periods.)

Here, we are comparing a path length difference and a wavelength and looking for some relationship. The uncertainties on these two values help us to have confidence in whatever relationship we think we see. (Repeated measurements under different conditions can help build that confidence, but we're stopping after measuring for one maximum and one minimum today.) So, if your wavelength is $\lambda = 101 \pm 5$ nm and your path length difference is $\Delta= 24 \pm 2$ nm, you would say your measurements are consistent with $\Delta = \lambda/4$ (and not consistent with, for example, $\Delta = \lambda/3$ or $\Delta = \lambda/2$).

In the simulation above, we considered what happened when we had two sources which produced in-phase waves of identical frequency. You observed interference in that the wave intensity changed depending on the position; in some locations the two waves constructively added to form an intensity maximum, while in others they destructively cancelled each other out. Importantly, in this case the intensity at a given position was *constant in time*. A spot was, for example, *always* an intensity maximum or *always* an intensity minimum.

This is, however, not the only way we can get a phase difference between two waves and interference. We can use two wave sources which have *different frequencies*. Now, at a given position, the wave amplitude will increase when the phase difference between the two waves is small (that is, when the two waves nearly overlap) and will decrease when the phase difference is close to 180° (that is, when the crest of one wave is nearly in line with the trough of the other). Whether the waves add constructively or destructively will change continuously; a given point will cycle from fully constructive to fully destructive and back.

This slow variation in the amplitude of the sum of the two frequencies is called *beating*. The frequency of the beat is given by the difference between the two frequencies from each source. Figure 1 shows this amplitude variation created when two waves of similar frequency interfere at a point.

Figure 1: Two waves of identical amplitude, but slightly different frequency add to produce a wave whose amplitude varies with time with a frequency equal to the frequency difference between the two waves.(Image borrowed from oPhysics.) Click image to see animation. |

In order to *experience* beats, we have produced a few sound files that mix pure tones of equal volume but different frequency. Listen to these files and see if you can hear the beat frequencies.

Sound File | Frequency 1 | Frequency 2 | Beat Frequency |
---|---|---|---|

800 Hz | 700 Hz | 100 Hz | |

800 Hz | 750 Hz | 50 Hz | |

800 Hz | 790 Hz | 10 Hz | |

800 Hz | 799 Hz | 1 Hz | |

800 Hz | 800 Hz | 0 Hz |

Now, take a listen to the two *mystery* tones below.

Mystery Tone 1 | Mystery Tone 2 |
---|---|

Using this online tone generator, you can play the mystery tone and a tone of known frequency at the same time.

- By listening for when the beat frequency goes to zero, can you identify the frequencies of the two mystery tones?
- Are either (or both) the tones an integer frequency, or does your mystery tone seem to fall in between the values available in the tone generator? Can you estimate the mystery frequency value anyway?

We can now use an adjustable tone generator that will produce different frequencies in each channel of your speakers (or headphones) – one tone in the left channel and one in the right channel:

**oPhysics Interactive Beats Simulation**: https://ophysics.com/waves10.html

This part only works if you can listen to each channel separately. It's easiest to do with headphones, but it can also be done if you have separate (and small) left and right speakers.

If you don't have access to speakers or headphones like this, then please read through the following section anyway, but write “I do not have access to the appropriate equipment for this part of the experiment and cannot complete this section” on your lab report template.

Choose two frequencies that are different (but close) at about 200 Hz. Place **both** the left and right headphones close to one of your ears. Adjust one of the frequencies until beats are clearly audible.

- Describe what you hear.
- Based on the frequencies you chose, what beat frequency should you hear? Is that what you detect?

With both phones at one ear, the mechanism for hearing beats is easily understood: while the two waves are in phase, the air pressure driving the ear drum is larger and the sound is perceived as louder. While out of phase, the pressure is less and is perceived as quieter.

Next, put on the headphones so that one sound plays in one ear and one sound plays in the your other. In this mode, each ear drum is presented with a single frequency and should, therefore, not experience beats.

Without changing the frequencies used earlier, place one phone on each ear and listen for beats.

- What do you hear?

Repeat the above exercise using frequencies of about 500 Hz and 1000 Hz.

- Can you still hear beats at this higher frequency?

The physiology involved here may be summarized as follows: Each eardrum is forced to vibrate by the small pressure changes caused by the vibration of its headphone. The eardrum is mechanically coupled through small bones to another membrane at the entrance to the cochlea, a fluid-filled spiral tube. Inside the cochlea is a membrane containing hair cells which are set in motion by the vibration of the fluid. The hair cells closest to the cochlear entrance are sensitive to high frequencies, while those farther from the entrance are sensitive to lower frequencies. The hair cells convert the mechanical vibrations to electrical signals. The signals from the left and right ears are *mixed* in each of two sets of neurons in the brain stem. One set of neurons is sensitive to high frequencies (typically kHz.) and detects *intensity*. At high frequencies, the sound shadow cast by the head gives rise to differences in intensity from left to right and is used to judge direction of the sound source. The other set of neurons, sensitive to lower frequencies (typically 200 Hz.), detects *phase* or time differences to judge the direction of the sound source. The detection of beats with different frequencies sent to the left and right ears depends on the functioning of the lower frequency, phase-sensing neurons.

It has been observed that some people can hear beats in the two-ear mode while others cannot.

- According to the above model the ability to hear beats in this mode should be better at low frequencies. Is this the case for you?

A very well-known device for observing interference between light waves is the Michelson interferometer shown in Fig. 2.

Its operation is as follows: A wave coming from the source impinges on the half-reflecting mirror. Half is reflected to Mirror 1, while the remaining half continues to Mirror 2. These two mirrors in turn reflect the wave back to the half-reflecting mirror. (The rays are shown displaced so you can trace them). The wave reflected from Mirror 2 is now reflected by the half-reflecting mirror, while that from Mirror 1 passes through the half-reflecting mirror, also to the detector. Thus, the original beam was split into two equal parts, traveled different paths, and then recombined.

The condition for interference is now the same as for the waves from two source, i.e., one gets constructive interference whenever the path differences are integer multiples of the wavelength. The path difference is changed by displacing a mirror along the light path. (Note that when a mirror is displaced the path length is changed by twice the displacement of the mirror.)

The interferometer we are going to use here is built for microwaves (a type of light), which have wavelengths that are on a scale easily measured by a regular ruler. (Visible light has a much smaller wavelength – on the order of a few hundred nanometers, or ~ $10^-7$ m – and a radio wave has a much longer wavelength – on the order of 10 m.

Watch the following videos and answer the questions that follow.

Also, here are a few clarifying points for things seen in the video:

- The length scale shown in the video has units of
**centimeters**. - The intensity meter sometimes shows a
*negative*deflection; this is just an artifact of the way the meter works. A big deflection to positive value means**greater**measured light intensity, but a big deflection to or below zero means**low**or**no**light intensity. - The beamsplitter looks like a plain piece of cardboard, but the side facing away from the camera has a thin coating of material which is designed to
**partly reflect**(and partly transmit) light of this wavelength.

- As the reflector moves along one arm, the intensity measured at the detector changes. Why? (
*Don't worry about the values or units on the meter; the scale is arbitrary. Just focus on qualitative changes.*) - Based on the positions where you see the signal go to a maximum (or a minimum), determine the wavelength of the microwave.
- Estimate the uncertainty in the microwave wavelength (and describe how you made that estimate).

Don't forget that the light traveling through each arm of the interferometer has to travel **towards** the reflector and then **back** to the beamsplitter. So, if the mirror is a distance x from the beamsplitter, the light travels a distance 2x along the arm by the time it goes there and back. Think about why this is important in your calculation.

The manufacturer says that their microwave transmitter produces light with a frequency of 10.5 GHz (or $10.5 \times 10^9$ Hz). They do not provide an uncertainty on that value.

- Is your measurement of the wavelength consistent with what you'd expect based on this frequency?

Remember to submit your lab notebook before your group meeting!

At the group meeting, you will discuss your findings for the three different instances of interference explored above. Be prepared to share your results and observations.

NOTEBOOK: You may take notes during your meeting if you find it useful, but it is not required. Part 2 will build on the physics explored in this first part, but it will explore that physics with completely new data and systems.

To help us start to understand diffraction, let's look at this video from *Veritasium* about a very curious phenomenon called “Arago's spot”. Observation of this non-intuitive phenomenon helped build the evidence for light's behavior as a wave.

The unusual patterns predicted by diffraction theory – like Arago's spot – were difficult to see using the technology of the 1800s (i.e., candles, slits, and screens.) Today, though, with the use of coherent laser beams and CCD camera detectors, it has become much easier to see these very curious – and pretty! – diffraction patterns, and to see the wave nature of light with our own eyes.

A simple setup is shown in Fig. 3. A helium-neon laser produces red light which passes through a diverging lens to spread the beam over a wide area (to cover the CCD of the TV camera and to reduce the beam intensity). There are several objects provided for the beam to pass through (e.g., a sewing needle, a razor blade edge, and some ink spots on a glass slide) which are shown in more detail in Fig. 1(b).

As we move the different objects into the beam to cast a shadow onto the TV camera, different diffraction patterns appear on the monitor. These images are shown in Fig. 4.

Look over the five images in Fig. 4 above.

- For
**one**of the images, describe the pattern you see and (in one or two sentences) explain how the symmetry (or asymmetry) of the object leads to the pattern.

Now let's make some quantitative measurements by looking at diffraction through a single slit.

In the interference portion of the lab, we saw that light from two point sources would interfere at different positions in space due to the superposition of the waves from each source.

Figure 5 represents the wave disturbance originating from two coherent point sources, S1 and S2, separated by a distance $d$. If we place a detector at some point, the intensity at the point is given by the superposition principle. The intensity 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. $\Delta r = n\lambda$. Since the path length difference for these two sources is $\Delta r = d\sin\theta$ (where $\theta$ is defined in the figure), the angles at which the intensity will be a maximum, denoted by $\theta_{\textrm{max}}$, obey the following relation:

$n\lambda = d\sin\theta_{\textrm{max}}$ (for $n=0,1,2,\dots$). | (1) |

The intensity observed at the detector will be a *minimum* if the two sources are out of phase, i.e., $\Delta r = (n+1/2)\lambda$. Therefore, denoting angles where we expect an intensity minimum by $\theta_{\textrm{min}}$, we have

$(n+1/2)\lambda = d\sin \theta_{\textrm{min}}$ (for $n=0,1,2,\dots$). | (2) |

Now, consider an extended source, e.g., a slit, having width a as shown in Fig. 6. 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.

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

$\Delta r = \dfrac{a}{2}\sin\theta_{\textrm{min}} = \dfrac{n\lambda}{2}$ (for $n=1,2,3,\dots$). | (3) |

(Note that $n = 0$ is not included.) Thus,

$a\sin \theta_{\textrm{min}} = nλ$ (for $n=1,2,3,\dots$). | (4) |

The maxima will occur approximately half way between the minima. A single slit diffraction pattern is shown in Fig. 7.

To create diffraction patterns, we will direct a laser through a specially-designed slit holder and onto a white screen. (See Fig. 8) In front of the screen, we mount 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.

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.)

Let us try to determine the wavelength of the laser from the diffraction pattern.

From the data provided in Fig. 7 for the single slit diffraction, fill out the following table 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 uncertainties. It is possible to estimate uncertainties on all the measured quantities and to propagate them through to a final result. It will be a bit of a complicated formula to derive, but once you have it, it should not be too hard to use. (You may want to look over this page for a reminder of the general propagation formula using partial derivatives.)

If you are unable to use the propagation formula, you may use the following approximation: $\dfrac{\delta \lambda}{\lambda} = \dfrac{\delta x_{\textrm{min,left}}}{x_{min}}$.

For example, if you think you can estimate the position of the first minimum with $\delta x_{\textrm{min,left}} = 0.20$ cm and the distance from the center is $x_{\textrm{min}} = 4.00$ cm, then the fractional uncertainty in your wavelength will be $\delta \lambda/\lambda = \delta x_{\textrm{min,left}}/x_{\textrm{min}} = (0.20)/4.00 = 0.05 = 5\%$.

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?

We saw above that interference from two point sources separated by distance $d$ produced maxima on a detector at angles given by

$n\lambda = d\sin \theta_{\textrm{max}}$ (for $n=0,1,2,\dots$). | (5) |

When we change from two point sources to two narrow slits, we now have a combination of the interference due to two sources and the diffraction due to the non-zero slit width; the result is an intensity pattern similar to that shown in Fig. 10. Note that the two slit pattern of Fig. 10 would lie entirely beneath the single slit pattern of Fig. 7, showing that the larger overall shape (i.e. the envelope) of the two slit pattern is still dictated by the widths of the individual slits.

We can, in fact, show this experimentally as illustrated in Fig. 11. The single slit diffraction pattern for slit width a = 0.08 mm is shown again in Fig. 11(a). The diffraction pattern for light through two slits of width $a$ = 0.08 mm and separation $d$ = 0.25 mm is shown in Fig. 11(b). The overlap of the two is shown in Fig. 11©, with the double-slit pattern clearly falling completely under the envelope formed by the single-slit pattern.

As we continue to add equally-spaced slits – 3 slits, 4 slits, etc. – the position of these maxima do not change, but they become progressively narrower. In the limit of many slits, we effectively find destructive interference everywhere except right at the maximum angle. Such a device is called a **diffraction grating**, and it might have hundreds or thousands of slits (or more commonly, shallow grooves which act the same way) and therefore can achieve diffraction spectra with incredibly sharp peaks.

The apparatus we will use is shown in Fig. 12. We mount a laser – either red, green or blue – and direct it at a diffraction grating that has 600 grooves/mm. The beam then continues onto a white ruler, when the interference maxima can be measured.

Placing our ruler a distance $L = 9.8 \pm 0.2$ cm from the diffraction grating, we take the pictures of the ruler shown in Fig. 13.

Again, let's use our diffraction pattern to estimate the wavelength of the light.

- If the diffraction grating has 600 grooves/mm, what is the spacing between grooves, $d$?
- Based on these diffraction patterns, which color of light has the longest wavelength? Which has the shortest?
- Using this spacing, determine the wavelength of one of the lasers from estimates of the maxima positions on the rulers. Use the data from as many visible diffraction maxima as possible to make your estimate.
- Using a similar uncertainty approximation to the one we had above, estimate the wavelength uncertainty as $\delta \lambda/\lambda = \delta x_{\textrm{max,left}}/x_{\textrm{max}}$. (You can calculate this fractional uncertainty using the $n = 1$ maximum measurement only.)
- How does your estimate compare to the known wavelength: $\lambda_a$ = 652 nm, $\lambda_b$ = 532 nm, and $\lambda_c$ = 405 nm? (We won't tell you which laser is a, b, or c. You have to decide which wavelength your measurement should be compared to.)

Again, submit your (updated) notebook before the meeting.

Be prepared to discuss your results from the diffraction exercises above with the group.

After your second meeting, you will again need to write up your summary and your conclusions. Include any data tables, plots, etc. from the experiment or discussions as necessary in order to show how your data support your conclusions.

This part doesn't need to be long; one or two pages should be sufficient. What is important, however, is that your writing should be complete and meaningful. Address both the qualitative and quantitative aspects of the experiment, and make sure you cover all the “take-away” topics in enough depth. Don't include throw-away statements like “Looks good” or “Agrees pretty well.” Instead, try to be precise.

Remember… your goal is not to discover some “correct” answer. In fact, approaching any experiment with that mind set is the exact wrong thing to do. You must always strive to reach conclusions which are supported by your data, regardless of what you think the “right” answer should be. Never, under any circumstances should you state a conclusion which is contradicted by the data. Stating that the results of your experiment are inconclusive, or do not agree with theoretical predictions is completely acceptable if that is what your data indicate. Trying to shoehorn your data into agree with some preconceived expectation when you cannot support that claim is fraudulent.

**REMINDER**: Your report is due 48 hours after the end of your meeting. Submit a single PDF on Canvas.

Below are some resources for TAs to use. If you're a student… you can look, but you don't need to go through or understand anything here (unless your TA asks you to explore any of these things during your meeting.)

**Spectrometer**

A diffraction grating is the key element in a device known as a **spectrometer**, shown in Fig. 14. A spectrometer can be used to identify the wavelength(s) of light from a source… be it a monochromatic (i.e. single color or wavelength) source like the lasers above, or a more complex source which contains many wavelengths. (Or even a continuum!)

Thinking back to the last part of the lab – light from the three different-colored lasers through the diffraction grating projected onto a ruler – try to understand how the spectrometer acts a bit differently. Whereas above we saw the whole diffraction pattern displayed at once, here we will only see a little bit at a time depending on what angle the telescope is at. (When you are at an angle where a diffraction maximum appears, you will see light. When you are at an different angle, you will see dark.)

A spectrometer is a *very precise* instrument. If you know the diffraction grating spacing and you measure the angle to high precision, you can calculate the wavelength of light to high precision.

**The “experiment”**

We looked at what happens when you point a laser at a slit or diffraction grating, but a laser is a monochromatic source. (That is, it emits a single color or equivalently, a single wavelength.) What kind of pattern (if any) do you think you would you see for the following light sources shining through a diffraction grating and why?

- A neon sign
- A light bulb
- The sun

When scientists in the early 1800's looked at the sun with a spectrometer, they noticed some unexpected features. They saw many colors of light, but they also some some dark bands in the spectrum (which meant that some colors were *missing*). Today, we call these bands the **Fraunhofer lines** (named for Joseph von Fraunhofer who studied them in 1814, even though William Hyde Wollaston noticed them first in 1802.) Figure 15 shows the most prominent of these lines.

Figure 15: Solar spectrum with Fraunhofer lines indicated. (Source: Wikipedia) |

In 1859, several scientists noticed that when they pointed their spectrometers at gas lamps made from certain elements (like sodium or mercury), the light they emitted was not continuous at all, but instead was very discreet (and sometimes even as monochromatic as a modern-day laser). And in fact, some of the *emission wavelengths* from these elements matched up with the missing wavelengths in the solar spectrum. See, for example, the hydrogen emission lines shown in Fig. 16.

So…

- What's going on (both here on Earth and on/in the Sun)?
- If an element is found to correspond to one of the missing colors in the solar spectrum, what does that mean?
- How can we design experiments to look at the emission and absorption spectra here on earth?