Grating Challenge

• Is the pattern of colors repeatable? That is, does red always appear in the same location relative to blue (for example), independent of which kind of light you look at, or is it more complicated than this?
• How does the pattern vary as the grating is moved closer to or further from the light source?
• How does the pattern vary the grating is moved closer to or further from your eye?
• What happens to the pattern as you tilt the grating relative to the plane of your face?
• Is there a proper orientation to the grating? That is, what happens when the grating is rotated? What happens when the grating is flipped over and you look through the back of the slide?

The bands you will be looking for are fairly faint, so you will likely want to set this up in a dark space. We used the SUNY Cortland planetarium for this project since it is easier to get good photos in a large space like this. Our light source is the flashlight on a cell phone, which is leaning against a brick (at the bottom of the picture shown here). Light passes from the source through the grating (the white card mounted on the arm that is connected to the stand) and onto an observation screen, which was placed 20 cm from the grating. We taped a ruler across the image so that we could directly measure the band positions from the photos, but you can just mark the color bands with a pencil on a paper and measure afterward.

You can make a simple and effective DIY spectrometer with a cardboard box. Note that the colors you will be looking for are really faint, so you will need to make your spectrometer fairly light-proof. In this setup, the viewer looks in toward the observation screen (the back of the box) from the angled viewing port on the top. We added a light shield between the viewing port and the light source to block stray light from the source, which can be significant enough to cause one's eye to become less sensitive to the dark and therefore harder to see the faint diffraction patterns that we are looking for.

Inside the spectrometer we drew a central line and a measurement axis labeled with centimeter increments. The black rectangle in the middle is there to try to absorb some of the white light of the central beam, which can reflect off of the back of the box and cause a significant amount of light pollution in the interior, making it harder to see the faint diffraction pattern. Maybe you can devise a better way to manage this unwanted light.

You are seeing a broad white pattern because the wide area of the grating is admitting too much light and your diffraction pattern is getting washed out by the broad background light. To make this work you have to limit the region through which light passes.

An object that limits the amount or extent of light entering an optical system is called an aperture. You can create an simple and effective linear aperture by attaching two sticky notes to the grating. Note that the direction of the aperture matters. It might be interesting to see what happens with the aperture oriented perpendicular to that shown here.

For this, we will need to use some trigonometry. Or a protractor if you have one, but getting a good estimate of the angle might not be so easy given that your protractor is on one plane and the observation screen is on another plane.

To do this using trig, we also need to know the distance from the grating to the observation screen. Calling the position of a particular color x and the distance from the grating to the screen L, we have tan(θ) = x/L. You can now solve this equation for θ by using the inverse tangent function (arctan). Alternatively, since we know that we need sin(θ), we can use the Pythagorean theorem to get the hypotenuse of the triangle, which allows us to develop an equation for sin(θ).

For small angles, we can use the approximation that sin(θ) ≈ x/L. Of course, this is not perfect, but the error from this approximation is likely quite a bit less than the uncertainty from our experimental setup, so it doesn't make sense to worry about it.

The picture of our setup, using the grating with a linear aperture, shows nice bands of color to either side of the white part in the middle. We added a black strip of paper down the center so that the photo did not saturate due to the bright light reflected from this region, but you won't need to do that. The center of the diffraction pattern is close to the 14 cm mark. We see that the nearest edge of the violet band is about 4.0 cm from the center and the far edge of the red band is about 7.5 cm from the center. The grating was placed 20 cm from the observation plane.

violet light: d ≈ (400 nm)(20 cm)/(4.0 cm) ≈ 2.0 μm

red light: d ≈ (750 nm)(20 cm)/(7.5 cm) ≈ 2.0 μm

Alternatively, had we used the exact formula for sin(θ) we would have calculated 2.0 μm and 2.1 μm, respectively. Note that μm is a micrometer, which is equal to 1 /1000th of a mm. Expressed as a number of lines per unit length, 2.0 μm between lines is equivalent to 500 lines/mm, 5000 lines/cm, or 1970 lines/inch.

The diffraction pattern of the filament shows the same structure as the filament when viewed from the end.

The bulb viewed from the side has all legs of the filament aligned, or nearly aligned, which appears to us as a vertical line in the center of this image. Notice, however, that the diffraction pattern of the filament viewed from this angle looks more like a "C" than a vertical line.

The pattern generated by the diffraction grating is a type of hologram. That is, it preserves something of the 3D structure of the object being imaged. In the case of the filament viewed from the end of the bulb, the entire filament is at the same distance away from the grating, so the pattern produced essentially just replicates what we see. However, for the filament viewed from the side, the filament is an extended object that exists at different distances from the grating. This means that light from the different parts of the filaments enters the grating at different angle and therefore produces different diffraction pattern, which reveals to us the hidden structure of the filament that we cannot see by directly looking at it.

Our intuition correctly tells us that the filament will glow brighter as we increase the voltage across it. This is rather subtle, as it is not really the voltage that determines the brightness, at least not fundamentally (though it is our easiest control knob to see a variation, practically speaking). The real story here is that as the voltage is increased the current that flows through the filament increases, and with more current comes more heating of the filament (higher temperatures). The resistance of the wire is better thought of as a function of temperature, which increases with temperature, which happens to occur as voltage is increased, though the increased voltage is not the direct cause of the resistance increase.

We can go a step further and calculate the resistive power delivered to the filament as V²/R, where V is the voltage across the bulb (from the variac) and R is the resistance of the filament. For constant resistance, doubling the voltage will result in an increase in power by a factor of 4. There is a wrinkle here, however. As the filament heats, its resistance increases. But how exactly it increases is unclear: is it linear, quadratic, exponential, something else? Turns out that is a somewhat complicated story, and we need to go into the details here other than to acknowledge that resistance increases. The plot below is of data taken in a lab at SUNY Cortland using the bulb and variac system shown above. The resistance is not directly measured but is inferred from current and voltage measurements.

The electrical power dissipated in the resistor can be calculated as the product of the RMS voltage and the RMS current, that is P=IV. Using Ohm's Law (which we can do now that we know how the resistance changes with voltage), we can use I = V/R to get P = V²/R. Because resistance increases with the applied voltage, we expect that the power versus voltage curve will be less than quadratic. The graph below shows exactly this, with the gray line showing the curve we would expect if the resistance were constant (at its lowest value from the graph above), while the blue points show the calculated resistive power using P=IV.

A very reasonable first place to start would be to consider the relationship between the brightness of an object and its temperature, which for an object like the filament of a bulb must be related to the electrical power that flows into it. Clearly, the filament at 100 Volts is quite a bit brighter than the filament at 50 Volts, and that even with the exposure time reduced by a factor of 4.

There is another way in which the pattern of light changes with temperature. As an object gets hotter, the peak of its radiation spectrum shifts to smaller wavelengths. We can see that in the images above where the 50 V case shows only red and green light, whereas the 100 V case shows a full spectrum of visible light with just a faint hint of violet light. This patter of radiation is well known and is called "blackbody" radiation.

The theoretical blackbody spectra for two objects: one at a temperature of 2000 K (1727 C, 3140 F) and the other at a temperature of 2700 K (2327 C, 4220 F). Note that the plot for the cooler object was increased by a factor of 3 to better compare the shapes of the two curves. The thin blue line lower in the figure shows what this curve would look like if its scale were not increased. The solid vertical lines indicate the limits of the visible range, and the dotted vertical lines span the range of green colors within the visible spectrum as a point of reference.

As a last point, it is interesting that so much of the radiated power occurs at wavelengths larger than the visible spectrum. All of this is infrared light that is invisible to our eye, but which we can feel as radiant heat. Whenever we use an incandescent light bulb, we have to produce this invisible light. For the purposes of creating light to see by, this is terribly inefficient. If we wanted to use a light bulb to also generate heat, this isn't such a bad solution.

Of course! The key to comparing the electrical and spectral data is recognizing that essentially all of the electrical power dissipated in the filament must leave as radiant energy (the only other pathway would be conductive losses through surrounding material, but that is very small).

Noting that the 100 V picture had very faint traces of violet, we see that this is represented pictorially in the graphs of blackbody radiation shown above by relatively small value of the orange curve near 400 nm. The 50 V case showed signs of green colors, though quite a bit fainter than the reds, and no discernible traces of blue. This is well represented by the blue curve in the graph above since it has small amplitude in the green range and effectively goes to zero for shorter wavelengths. This is all very rudimentary, but from this we could estimate that the temperatures of the filaments are about 2000 K and 2600 K, respectively. These theoretical models for blackbody radiation can also give us a measure of the total radiant power, which is about 2.9 times larger at 2600 K compared to 2000 K.

This number can be compared against the measurements of the electrical power, which show that the power to the filament is about 3.1 times larger at 100 V compared to 50 V. Is this a success or a failure? That is up to you to decide. On the one hand, these two numbers are pretty far off (2.9 compared to 3.1). On the other hand, this is actually pretty good since the spectral analysis was very speculative and we more-or-less just guessed at a few temperatures when making these plots. What this shows is that we got close to the right answer and that there is clearly lots of room for improvement.