**by
Dr. Didar Islam,
Central Michigan Uuniversity Professor of
Physics**

**GOAL: To use the concepts of superposition, interference, and diffraction to identify small particle size (on the order of 1mm).**

__General Information __

When light passing through a narrow slit is allowed to illuminate a screen, bright and dark regions appear. The intensity of the illumination of each point on the screen, bright or dark, is proportional to the *square of the amplitude* of the light wave incident at that point, and not just the amplitude. How are these bright and dark regions produced?

The answer to the question above is not really complex and will become clear once we understand some basic concepts like diffraction, the superposition principle and interference. DIFFRACTION occurs whenever a wave front travels past the edge of an obstruction in its path or through a small aperture. As the wave front passes the edge or aperture, secondary spherical waves emerge from the small region near the edge or aperture. In Figure 1a, plane waves are incident upon an obstruction. The parallel lines represent crests on the wave front. The distance between each crest is a wavelength. In the diagram, can you identify where the troughs are located? Draw the positions of the trough using colored pencils or pens. To visualize what happens in diffraction, draw the position of the crest and trough of spherical waves emerging from the edge. Remember that the wavelength of the emerging wave front is the same as the incident wave.

Figure 1b shows a plane wave front incident on a single slit whose width is of the order of the wavelength of the incident wave front. Carefully draw the emerging diffracted spherical wave fronts from near each edge. Also, draw the positions of the troughs using colored pencils or pens.

If you have successfully drawn the diffracted spherical waves you will notice that the crests and troughs from the two sets of spherical waves you have drawn overlap at every point beyond the screen. The amplitudes of the two secondary spherical waves add algebraically at each point to give new amplitude at each point. Note that the amplitudes of the two waves add up but not the intensities. This is called the SUPERPOSITION PRINCIPLE. The intensity of light at any point is the square of this new amplitude at that point. The two spherical waves are now said to INTERFERE with each other because the intensity depends on how the amplitudes add up at that point.

If you look carefully at your drawing, you will notice two extreme ways in which the waves interfere. At certain points, a crest adds to another crest (or a trough adds to another trough) resulting in a larger amplitude (positive or negative). The light intensity is high at these points since the intensity is the square of the amplitude. This is called CONSTRUCTIVE INTERFERENCE. At certain other points, a crest adds to a trough resulting in a smaller (and sometimes nearly zero) amplitude. The light intensity at these points is almost zero. This is called DESTRUCTIVE INTERFERENCE. All other interference effects are between these two extremes.

Figure 1a Plane wave incident on a single edge.

Figure 1b Plane wave incident on a single slit.

A direct consequence of the superposition principle is that the observed intensity of diffracted light falling on a screen has bright (constructive interference) and dark (destructive interference) regions or bands. You can use the figure you have drawn around Figure 1.b to understand, in principle, why dark and bright fringes occur on the screen. Identify where on the screen the bright and dark fringes will appear as a result of constructive and destructive interference of the two spherical waves emerging from the two edges?

In an actual single-slit diffraction experiment, the angular positions of the bright and dark fringes will be somewhat different from what may be expected from your drawing in figure 1a. The actual interference pattern arises due to the superposition of spherical waves from all points on the aperture between the edges, and not just the edges. Clearly, this will make your drawing quite cluttered, so you need not attempt to draw all these spherical waves. However, you can try and visualize on the diagram the effect of the superposition of all those waves. The relative position and distribution of these bright and dark regions are predictable and depends upon the wavelength of the light used (l ), the width (a) of the slit and the distance (L) of the screen from the diffracting slits.

This is represented in Figure 2 below.

Figure 2. Location of first pair of dark fringes. (See__ Physics: Principles and Problems,__ by Zitzewitz and Neff, Glencoe, 1995, page 397.)

Similar results can be obtained if a thin wire or hair is substituted for the slit between the edges in Figure 1b.

Figure 3 below shows a schematic diagram of a laser particle detector. The arrangement depicts the laser radiation passing through a region of small particles contained in the Sample Cell. The lens focuses the light on the analyzer at the far right which analyzes the diffraction pattern, giving particle size.

Figure 3. Schematic diagram of a laser particle size analyzer.

For a single slit (whose width is on the order of the wavelength of the light)or for a single obstruction, such as a wire (whose diameter is on the order of the wavelength of the light) the positions of the dark regions can be predicted by:

m = 1,2,3...Equation 1

The variables are:

l wavelength

a slit width

L slit to screen distance

q angular position of dark fringe

x Distance, on the screen, form center of central bright fringe to the dark fringe

m order number of diffraction. Starting from the center of the screen in either direction, m = 1,2,3... for the first, second, and third order.

By Equation 1, the smallest q corresponds to m = 1.

Since q is usually very small we can use the fact that for small q, sin q = x/L approximately. Equation 1 can then be written as:

m = 1,2,3...Equation 2

Equation 2 can be used to calculate slit or obstruction width once n and l have been determined from experimental measurements. The first dark region is usually well defined and therefore is used in calculating the width of the slit or obstruction.

The following information will not be used in this lesson but is worthwhile to keep in mind in identifying particle sizes. For a narrow slit or obstruction, Equation 1 gives:

Equation 3

where q _{1 }is called the half-width of the central maxima

When the slit or obstruction is circular (or nearly so), the distribution of the diffracted spherical waves reaching the screen is significantly different from what has been considered above for a narrow slit. As a result of the new geometry, instead of Equation 3 the correct formula for a spherical aperture or obstruction is:

Equation 4

Equation 4 can be used to determine the size of unknown circular apertures.

ACTIVITY ONE: WIDTH OF A HUMAN HAIR

**GOAL: To determine the width of a human hair by using diffraction of laser light. **

Part I.

The diffraction pattern below is for a human hair of unknown diameter. The diffraction pattern was recorded under the following physical conditions:

l 633 nm

L ft

The scale factor is 2:5. This means a 2mm measurement on the diagram converts to an actual measurement of 5 mm.

Pattern 1. Diffraction Pattern created by a human hair.

In Pattern 1, measure the separation between the centers of the first dark regions on either side of the central bright region. From this measurement, determine the variable x. Remember to use the scale factor 2:5. Using this, and the previous given variables, calculate the average width w of the hair. Repeat the measurements for the other dark fringes. Use Equation 2 and the corresponding values of m. Record the data and results in a tabular form.

Part II.

Using a standard classroom laser, design an actual experiment where you calculate the size (diameter) of your hair. Compare your results with those of your classmates. Explain why there are variations in hair diameter. Perhaps we could ask the following questions.

1. Is this method accurate enough to determine the difference between the hair diameters of the students in your class? What is the accuracy of this experiment?

How many significant digits can be obtained in the results of this experiment?

ACTIVITY TWO: SIZE OF SMALL UNKNOWN OBJECTS

**GOAL: To determine the size and shape of an unknown small object using diffraction of light.**

It is possible to determine the shape and size of a small object like a tiny grain of sand or dust by using the method employed above. This method is widely employed in industry and by researchers to determine unknown particle sizes. It is also used to determine particle size distribution in an aggregate of particles.

Pattern 2 is the interference pattern for a small object of unknown size. The interference pattern was recorded under the following physical conditions.

l 633 nm

L 22.0 ft

Scale factor 2:5

Pattern 2. Diffraction pattern of an unknown small object.

1. Observe the distribution of the bright and dark fringes. Can you guess the shape of the object?

2. Notice that the interference fringes are prominent in two different directions, left-to-right (the x-axis) and up-and-down (the y-axis). Can you guess if the object is wider along the x-direction or the y-direction? First consider the x-axis fringes. Measure the separation between the first two dark regions on either side of the central maxima and record it as 2x_{1}**.** Now measure the separation between the next two dark regions on either side of the central bright fringe and record it as 2x_{2}**. **Record all data in a tabular form. Make sure you record the order number for each measurement.

3. Repeat the procedure for the y-axis fringes and record the data as 2y_{1} and 2y_{2}. Record the data in a separate table for this data.

4. From the given and collected data, for destructive interference in the horizontal (x-axis) and vertical (y-axis) directions for each order number m, calculate the width of the object along the x and y directions using Equation 2 and record the result in the data table. Now what can be said about the size and shape of the object?

5. Were your guesses in steps 1 and 2 correct?

Questions? Comments??