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| 1. |
(a) Show that the values of a at which intensity maxima for single-slit diffraction occur can be found exactly by differentiating Eq. 35-56 with respect to alpha and equating the result to zero, obtaining the condition tan alpha= alpha. To find values of a satisfying this relation, plot the curve y = tan alpha and the straight line y = alpha and then find their intersections, or use a calculator to find an appropriate value of alpha by trial and error. Next, from alpha = (m + 1 // 2)pi, determine the values of m associated with the maxima in the single-slit pattern. (These m values are not integers because secondary maxima do not lie exactly halfway between minima.) What are the (b) smallest alpha and (c) associated m, the (d) second smallest alpha and (e) associated m, and the (f) third smallest alpha and (g) associated m? |
| Answer» Solution :(b) 0, (C) -0.500, (d) 4.493 RAD, (E) 0.930, (F) 7.725 rad, (G) 1.96 | |