Name: Lab Partners: Date: Pre-Lab Assignment: Interference, Measuring Wavelengths, and Atomic Spectra (Due at the beginning of lab) Directions: Read over the lab handout and then answer the following questions about the procedures.. What is your prediction.?. What is your prediction.? 3. What is your answer to question.? 4. Why do we use a diffraction grating instead of the two slit interference pattern in order to determine wavelengths accurately? PHYS-04: Physics II Laboratory i
Name: Lab Partners: Date: Interference, Measuring Wavelengths, and Atomic Spectra Objectives To learn the conditions that lead to constructive interference of waves. To develop a relationship between angle and wavelength for constructive interference for light coming through two parallel slits, or multiple equally spaced slits. To observe the distinction between the continuous spectrum emitted by a hot object and the line spectrum emitted by individual atoms. To discover how the relationship between wavelength and angle makes it possible to determine the wavelengths of light emitted by atoms. To determine the wavelengths of visible light emitted by hydrogen, and show that these wavelengths fit a simple relationship. Overview In the previous experiment we explored the behavior of light in conditions in which it can be treated as a ray, along paths. Although at the time of Newton there were no experiments that could conclusively show whether light was a ray or a wave, by the middle 800 s it had been shown that light is, in fact, a wave. The most important experimental observations that demonstrate this involve the phenomenon of interference. Interference effects are observable only when waves interact with objects near the size the wave length. For visible light, the wavelength is on the order of a half of a millionth of meter. The crucial property of waves that leads to interference is the fact that waves spread out in all directions from a source-including any place that a wave comes through an opening. This spread is greater if the light comes through a narrower slit. As a result of this spreading out light propagates from its source to other points by all possible paths. The intensity of light at a particular point depends upon whether the light waves arriving by way of different paths is in phase or out of phase. If one wave is 80 out of phase with another at a particular point, the waves cancel and we have no light arriving at that point. The simplest arrangement to analyze is Young s double slit experiment, where light passes through two narrow openings. If monochromatic, coherent light is incident from the left, the light which passes through the two slits will be in phase as it leaves the slits. Recall that constructive interference occurs when two rays are in phase. Constructive interference produces the maximum brightness. This will happen to the two rays directed at the center of the screen, as in Fig.. We will observe a bright spot at this location. However, waves coming out at an angle will have traveled different distances. The intensity of light coming away from the two openings depends on the difference in distance that the two waves have traveled-if the difference is one-half of a wavelength the peak of one wave coincides with the low point of the other. The two waves cancel each other, resulting in no light intensity at such an angle. This is called destructive interference, and is shown in Fig.. PHYS-04: Physics II Laboratory
Interference, Measuring Wavelengths, and Atomic Spectra v 0. Screen Figure : Light waves in phase because they have traveled the same distance. The crests of the two waves line up, so the two waves would combine to give a wave of large amplitude. Screen Figure : Light waves 80 out of phase because the wave passing through the lower slit travels one-half wavelength further. A dark spot would be observed on the screen at this angle. Note that the top ray travels s distance of 6.5λ while the bottom ray travels a distance of 6.75λ. At other angles the difference in distance may be equal to a wavelength, or wavelengths, or any other integer number of wavelengths. In this case the waves are in phase, and add up to give greater intensity, giving a bright region, as in the case of the waves that come out straight ahead. This is called constructive interference, and is illustrated in Fig. 3. The result is a series light and dark regions, called an interference pattern. If the light shines on a screen we will observe a series of light and dark bands, as in Fig. 4. The light regions correspond to angles at which constructive interference occurs; the dark regions correspond to destructive interference. The angles at which constructive interference occurs are related to the wavelength, and to the distance between the openings. Measuring these angles allows us to determine the wavelength of the light. Investigation : Determining wavelength from an interference pattern Look at Fig. 3, showing constructive interference. The difference in the distances traveled by the two waves is just one wavelength. We would call this first order constructive interference. PHYS-04: Physics II Laboratory
Interference, Measuring Wavelengths, and Atomic Spectra v 0. Screen Figure 3: Light waves in phase. At this angle the upper wave has traveled one wave length further so it is back in phase with the bottom wave. There would be maximum light intensity at this angle, as well as angles such that the upper wave travels an integer number of wavelengths further Figure 4: Interference pattern for two openings. If the difference in distance were two wavelengths, we would call it second order constructive interference, and so on. Prediction. Look at Fig. 3 closely enough to visualize how the angles at which first order constructive interference occurs for light of two different wavelengths compare. If red light has longer wavelength than green light. Will the angle for first order constructive interference with red light be larger or smaller than that for green light? Prediction. What happens to the angle for first order constructive interference if the distance between the slits is made smaller? PHYS-04: Physics II Laboratory 3
Interference, Measuring Wavelengths, and Atomic Spectra v 0. Question. Work out the answer to this question before coming to class. This is also question 3 of the Pre lab assignment. The Fig. 5 below shows the waves coming through two slits. To get constructive interference, the extra distance traveled by the upper wave must be nλ where n is an integer representing the order of constructive interference and λ is the wavelength. Use trigonometry to relate the angle, θ, the separation of the openings, d, the order, n, and the wavelength, λ. } nλ d θ θ Figure 5: The path difference between rays. Question. Is the equation you worked out consistent with your Predictions. and.? That is, does your equation imply that the angle θ must increase or decrease as the wavelength, λ, increases or decreases. Does it imply that θ must increase or decrease as the separation, d, increases or decreases? Activity.: Observing two-slit interference patterns The instructor will set up several lasers emitting light of different wavelengths to shine through a diaphragm that has a variety of openings, in it. This slit film is shown in Fig. 6. The sets of openings on the right side are double slits with a slit spacing that increases from the top to the bottom. Step : The instructor will shine red light from one of the lasers through the openings with the smallest separation (the openings at the top right in Fig. 6) onto a screen in the front of the lab. Then the light will be shown through the openings with the next smallest separation (second from the top). Observe the interference pattern for both cases Question.3 Which pair of openings gives the wider interference patterns? Is this consistent with your Prediction., relating the angle and the distance between the openings? PHYS-04: Physics II Laboratory 4
Interference, Measuring Wavelengths, and Atomic Spectra v 0. No. lines CAL width 3 space between lines 6 8 4 6 6 5 3 30 80 4 40 0 0.76 0.088 0.033 0.066 0.3 3 4 0 0.3 0.3 0.3 0.3 6 4 30 0.76 0.35 0.70.40 space between center of slits Figure 6: Slit film used for interference demonstration. Beside each element, the top digit refers to the number of lines, the middle digit to the width in points of the elements before 8X reduction, and the bottom digit to the space in points between lines. One point is 0.3546 millimeters The approximate distance in millimeters between the centers of the slits is shown under the patterns. Step : The instructor will shine light from the red laser, and light from a green laser through the openings with the smallest separation. Question.4 Which color gives the wider interference patterns? Is this consistent with your Prediction., relating the angle for constructive interference and the wavelength? Step 3: The distance from the diaphragm to the screen, and the displacement from the center of the interference pattern to the first bright area (the first order constructive interference) for both the red and the green light will be measured. You will be told the spacing of the two slits. Record this data in the table below. Distance to screen: L = m Slit spacing: d = m PHYS-04: Physics II Laboratory 5
Interference, Measuring Wavelengths, and Atomic Spectra v 0. Red light Green light Displacement: D Angle: θ Wavelength: λ Table : Red and Green Lasers Step 4: Note that the tangent of θ is related to the displacement and the distance to the screen, D. Calculate the angle for the first order constructive interference for each color, and then calculate the wavelengths using the formula you worked out in Question.. Enter the results of your calculations in Table. Question.5 Which has longer wavelength, red light or green light? In the next activity you will use a diffraction grating. The diffraction grating has many slits, not just two. The condition for the waves that have come through all the slits is just the same as for two slits: nλ = d sin θ. However, having many slits sharpens the contrast-now there will be a bright region only right at the angles given by this equation, instead of gradually fading from light to dark. Thus the angle, and therefore the wavelength, can be determined with much greater precision. Investigation : Continuous vs. Discrete Spectrum There are different types of light sources around the laboratory. Two lab stations have incandescent light bulbs, the ordinary type of light bulb in which a large number of atoms are collectively vibrating in the hot filament to produce light. The remaining stations have discharge tubes containing different gases at low pressure-hydrogen, helium, neon, and mercury. In the discharge tubes the light is emitted by individual atoms that have been excited by the electric discharge. Instead of projecting the light from these sources onto a screen as we did with the lasers, you will examine the light from these different light sources by looking at the sources through a diffraction grating. You will see light appearing to come not just directly from the light source, but from positions at several angles to either side of the source. Different colors of light appear at different angles, in accordance with the relationship between wavelength and angle. Activity.: Observing light from different sources. Since different students will have different light sources, some students will start with step, others with step. All students in a group should look at the light sources through the grating, so that they recognize the difference in the nature of the light from the incandescent bulb and the discharge tubes. PHYS-04: Physics II Laboratory 6
Interference, Measuring Wavelengths, and Atomic Spectra v 0. Step : Look at the incandescent bulb through the diffraction grating. Question. Qualitatively describe what you see. List the colors you see in order, starting with the color closest to the center. Include only the standard colors, red, blue, green, violet, yellow and orange in your list. Step : Look at the light from one of the discharge tubes through the diffraction grating. Question. Qualitatively describe what you see. Question.3 What is the distinction between what you have observed for the incandescent light and for the discharge tube? Why do we call one of these a continuous spectrum, and one a discrete spectrum? Which type of light source emits what you would consider a discrete spectrum? Comment: The distinction between a source that emits a continuous spectrum and one that emits a discrete spectrum is that the discrete spectrum is emitted by atoms or molecules acting independently of each other, whereas the continuous spectrum is emitted by atoms that are close together so that they interact strongly with each other. We refer to what you have observed with the discrete spectrum as spectral lines. Each of the lines for a given element corresponds to a particular wavelength. Each element has its own unique set of spectral lines. Step 3: Move around to different stations to observe the spectrum from each of the elements that you have not already looked at. By the time you are through, you should have seen the spectrum of hydrogen, helium, neon and mercury. Describe each of these qualitatively on the next page. PHYS-04: Physics II Laboratory 7
Interference, Measuring Wavelengths, and Atomic Spectra v 0. Hydrogen: Helium: Neon: Mercury: Investigation 3: Spectral Lines of Hydrogen Go to one of the lab stations that has a hydrogen or deuterium discharge tube to carry out this activity. The apparatus for this activity is arranged as in Fig. 7. Activity 3.: Measuring the Wavelengths of Light from Hydrogen Atoms. Step : Turn the discharge tube on if it isn t already, and place it so that it shines through the opening in the cardboard. Look through the grating at the light source. Note the position of the discharge tube as it appears on the meter stick and record it in Table. Also measure the distance from the grating to the meter stick and enter it in the table-you will need this value to calculate θ. Step : Look through the diffraction grating and identify the position of one of the lines on the meter stick. Record this position in the data table. PHYS-04: Physics II Laboratory 8
Interference, Measuring Wavelengths, and Atomic Spectra v 0. apparent position of spectral line slit displacement grating distance light source meter stick Figure 7: Apparatus for Measuring Spectral Lines. shown in Fig. 8. The appearance of the spectral lines is Step 3: Repeat your measurements for the other visible lines. You may only see one violet line. If you only see one, do not enter any data for the deep violet line in the table. Also, be careful that you are not recording data for a violet line in second order-if you go out farther than the red line in first order, you will see the whole sequence of lines repeated, corresponding to the second order constructive interference. R B V R BV VB R V B R m = m = m = 0 m = + m = + Figure 8: Three of the visible hydrogen spectral lines shown as a function of their relative displacement. The labels R, B and V denote red, blue and violet respectively. A fourth deep violet line is shown, but not labeled. Maxima occur when d sin θ = mλ, where m = 0, ±, ±,... Position of light source distance from grating to meter stick Spacing of the grating slits d = 530,000 m. =.89 06 m. Step 4: Calculate the displacement, which is equal to the difference of the position of the line and the position of the discharge tube for each spectral line, and enter the value in Table. Step 5: Calculate the wavelength for each spectral line. The series of spectral lines that you have observed are referred to as the Balmer series. Balmer had noticed that the wavelengths of these lines fits a pattern that can be described in PHYS-04: Physics II Laboratory 9
Interference, Measuring Wavelengths, and Atomic Spectra v 0. Tube position (m) Displacement from Center Angle θ Wavelength θ deep violet violet blue/green red Table : Hydrogen spectrum terms of integers. Specifically the series of wavelengths in meters are given by the formula ( =.0974 07 λ ) n where n = 3, 4, 5,... Question 3. Calculate the wavelengths for n = 3, 4, 5, 6 and 7. Do the wavelengths you measured for the three or four spectral lines you observed agree with any of these wavelengths? λ 3 = m λ 4 = m λ 5 = m λ 6 = m λ 7 = m Question 3. Since you were only able to see 3 or 4 lines, why are the others not visible? What part of the electromagnetic spectrum contains these wavelengths? If necessary, consult your textbook to find out about the parts of the electromagnetic spectrum. This laboratory exercise has been adapted from the references below. References PHYS-04: Physics II Laboratory 0