3. Combined two source interference and single slit diffraction. Consider a pair of slits in an opaque slide with center-to-center spacing d = 2.4 x 10 m and slit width D = 2 x 10 m. Light from a red laser of wavelength A = 7 x 10 m illuminates these slits and the resulting pattern is displayed on a screen L = 2 m beyond the slits. In this problem you will be plotting intensity patterns vs the coordinate y along the screen as shown below, where we will use the small angle approximation 0 ~ y/L. Figure A D y LThe combined intensity pattern is the product of the 2-slit interference and single slit diffraction: (kDy I1 = cos (kdy/ L) sin( 2L kDy (1) 2L where k = 27/1. (a) Make a plot of the above intensity pattern /1, using the parameters given. Make a plot with sufficiently high point density and over a broad enough range in y so that you get something that looks similar to the corresponding plot in Lecture 36 (it won't be identical, but should have the same qualitative features.) (b) Now suppose the individual slits are each replaced by a pair of slits with width D and spacing a = 3d/8 = 9 x 10-6 m as shown below. Figure B a =3d/8 V L 1 a =3d/8 These two pairs of double slits with spacing d offset by 3d/8 sort of mimics the situation for DNA, where there are two helices with a 3d/8 offset (see Lecture 37). Now the single-slit diffraction part of Equation 1 gets replaced by the two- slit interference x single slit diffraction of one of the individual pairs of slits, (kDy 12 = cos?(kdy/ L) cos(kay/ L)- sin ( 21 k Dy (2) 21 Make a plot of the above intensity pattern /2, using the parameters given. You may possibly want to focus in on the first few peaks of the pattern. Becausea = 3d/8 as in DNA you should see the 4th peak away from the center vanish just like in Rosalind Franklin's x-ray data (Photograph 51) for DNA! It is the interference between the scattered waves from the two helices (or pairs of slits here) that cause this cancellation. (c) Optional but interesting: Try plots with different values of a, say a = d/4 and a = d/2. You should see different peaks vanish corresponding to the examples to be shown in Lecture 37