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(a) Find the angle of the third diffraction minimum for 633 -nm light falling on a slit of width 20.0$\mu {m} .$ (b) What slit width would place this minimum at $85.0^{\circ} ?$ Explicitly show how you follow the steps in Problem-Solving Strategies for Wave Optics
A. $\theta=5.45^{\circ}$B. $D=1.91 \mu \mathrm{m}$
Physics 103
Chapter 27
Wave Optics
Simon Fraser University
Hope College
McMaster University
Lectures
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but a off the given problem. Linda Given we have these 633 nano meter converting this into a meter gives us tte come on, complying with a dental pile of minus nine. Love dies and meaning Beatrice, where, UM DVDs is a 20 times, tend to the power minus six meters. Then we can find it better for 1/3 diffraction. Minima, which is M, is equal to three. Well, then we can sort through these values in a formula which is a D sign. They don't is equal to m times, Linda. Then to substituting the values we solve for theta, the better we get here is 5.39 degrees in but be We had also find a D for the given theta, then again using the formula for car diffraction minimum, which is a tree times, um, 633 nano meter Converting into a meters that is multiplying. We tend to about minus nine, then dividing with given angle. Sign off the human angle, which is you know 0.99 gives us, um simplifying. This storm gives us a D is equal to 1.9 or six times 10 to the power minus six meters minus six meters
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