SOLUTIONS TO LAB SET 1 PROBLEMS

Diffraction from a Single Slit

[The intent here was to allow determination of slit width b just from measurements made from the image, given only the wavelength of the light used. However, in fact, the problem turns out to be quite difficult to solve, so I wish I had not assigned it as the first question in the first lab assignment.] The procedure is to realize that the geometry has to be determined, i.e., two of the following three quantities must be determined: {y displacement from central axis line, L separation between slit and screen, q_i angle from central axis to i^th fringe}. To determine two unknowns, two linearly independent equations must be obtained. If measurements are made of distances between two sets of corresponding dark fringe orders on both side of central axis, the results are not independent, because dark fringe spacings are multiples of each other. So have to measure bright fringes, which are not evenly spaced. Also, the measurement must be to the peak of the bright fringe, not what appears to be the center of the fringe. To do this, the image shown above must be converted in ImageTool to a grayscale image, and then a line profile taken of the horizontal axis through the center of the pattern. When this is done, it is clear that the peaks are noisy, so they must be smoothed by eye to pick the location of the brightest part of the fringe. It appears that the peaks are too noisy to allow enough accuracy to get reliable results. However, assume that the peak can be smoothed sufficiently to locate its peak accurately. The next step is to compare the measured distances from midpoint of the central bright spot to the peak of two bright fringes to the locations where the sinc²() function has its maximum values; these points can be taken from the included graph of sinc²(b). Since the sinc²() function has a maximum whenever its argument agrees with one of the peak values shown on the graph, and since b_i = (kbsinq_i)/2, where q_i = atan(y_i/L), the two data values can be obtained as follows. Let the two peak values of b taken from the graph be represented as b₁ and b₂, and let the corresponding measured distances taken from the image be y₁ and y₂. Then b₁ = p b sin(atan(y₁/L))/l and b₂ = p b sin(atan(y₂/L))/l are two equations in two unknowns (b, L). They can be solved (with considerable effort) to determine b.

Two-Slit Interference

Since the value of the angle to the third bright fringe is given as q₃ = 0.007718 rad, this problem is much easier. Ignoring the sequence of questions for a moment, the easiest measurement is to determine the slit spacing a by using the formula a sin q₃ = 3l which gives a = 0.246 mm. The simplest way to get the slit width b is to determine by inspection that a missing order occurs when the 8^th bright order coincides with the first diffraction order (dark). Therefore a/b = 8/1 or b = a/8 or b = 0.031 mm. To answer the final question concerning corroboration, a second method of determining b must be applied. To do this, use the given datum that q₃ = 0.007718 rad = y₃/L' for the third bright fringe. This gives the value of L' corresponding to whatever the magnification of the imaging system is. Measure the distance y₁ between the center spot of the image and the first dark fringe. Applying b sin (y₁/L') = b y₁/L' = l gives an independent determination of b, which can be compared to the value obtained by using the missing order approach.

Diffraction from a Square Aperture

This is really the same as Problem #1 and is solved the same way. Since the aperture is square, measurements can be taken either along the horizontal or the vertical axis.

Diffraction from a Circular Aperture

Skipping special case that occurs when b = 0, record the second and third given data values when the Bessel function J₁(b) = 0. Measure diameters y₁ and y₂ of the first two dark fringes. Set up two equations much like process in Problem 1.

Magnification of Monitor Display

Given the line density N (= 13,400 lines/inch) of the original grating, and the distance L (= 6.3 cm) between the grating and the Lab screen, determine the magnification of the image on the monitor display. This is rather easy, since some of the geometry is given. First invert the line density to get line spacing: a = 1 inch/13,400 lines = 7.463 10^-5 inch = 1.896 cm. Use first order bright spots with l = 633 nm, so a sin q₁ = l, which gives ay₁/L = l, where y₁ is the original lateral displacement of the first order bright from the central spot on the lab screen. Now use a ruler to determine y by measuring the distancebetween the +/- 1 orders on the display being used and dividing by 2. The magnification is then given by y/y₁.