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(a) If a single slit produces a first minimum at $14.5^{\circ}$ , at what angle is the second-order minimum? (b) What is the angle of the third-order minimum? (c) Is there a fourth-order minimum? (d) Use your answers to illustrate how the angular width of the central maximum is about twice the angular width of the next maximum (which is the angle between the first and second minima).
A. $30.1^{\circ}$B. $48.7^{\circ}$C. $\sin ^{-1}(1.0015)$D. 2$\theta_{1} \approx 2\left(\theta_{2}-\theta_{1}\right)$
Physics 103
Chapter 27
Wave Optics
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Lectures
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This is about a off a given problem. We had us to find the second order minimum angle which is indebted to we confined there to do by writing equations for forced order and second order round minimum and then simplifying for affected too. So d sign they him physical toe. I'm Linda from here. Sign that. I wonder why I ever signed that The two we get here is half. Then we console for a debtor to here. This will be signed in worse off two times off. Sign that a one given we have is a 14.5. So this uses 30.1 Dinis. But be that that the treaty will be signed in worse off warm three times. This is 34 then 30 minimum sign the one so substituting the values here three scientist The one we hear is 14.5. This gives us 48.70 piece similarly city part letter for fourth order and go there is signing worse off four times sine theta one which is a 14.5. Simplifying this expression we get that a four which is a ah this weekend one point 015 which is not possible. So it's no party two times off that one physical 2 to 14.5 logical to 29 degrees. So the two minus state of one is it called 2 30.2 or five minus 14.5 Excuse US 15.56 Thus, 29 Idris is approximately equal to two times 15.56 which is 31.1 degrees.
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