A single slit with a width of 0.100 mm is illuminated by light witha wavelength

Question

A single slit with a width of 0.100 mm is illuminated by light witha wavelength 610 nm.
(a) What is the angular location of the first diffractionminimum?
? rad
(b) What is the angular location of the second diffractionminimum?
? rad
(c) Diffraction maxima occur approximately halfway between adjacentdiffraction minima. Using your results from (a) and (b), calculatethe approximate angular location of the first diffraction peak tothe left or right of the central diffraction peak.
? rad
(d) To obtain the "exact" locations of diffraction maxima, valuesof alpha are required at which intensity maxima occur for a singleslit. An equation for these values of alpha are obtained bydifferentiating Eq. 36-5 with respect to alpha and equating theresult to zero. Two equations are obtained: (1) The first equation,sin / = 0, is the condition that must be satisfied when there is adiffraction minima. (2) The second equation of the form (sin )n1(cos )n2 = ()n3 is the condition that must be satisfied when thereis a diffraction maxima. What are the values of the integers n1, n2and n3, at least one of which is negative?
n1= ?
n2= ?
n3= ?
(e) It is not possible to exactly solve the equation fordiffraction maxima to obtain a general formula for the location ofall diffraction maxima; however, the values of alpha that satisfythis equation can be found graphically by plotting the curve y =(sin )n1 (cos )n2 and the curve y = ()n3 and determining where thetwo curves intersect. The smallest two values of alpha that satisfythe equation are alpha = 0 and alpha = 4.4934. The first solutionis exact, and the second and all others are only approximate. Usingthe solution alpha = 4.4934, solve for the angular location thetaof the diffraction peak adjacent to the central peak.
? rad
(f) Calculate the percent error between the actual value of thetacalculated in (e) and the approximate value calculated in (c).
????%

I have a and b need help with the rest. This is a modifiedversion of Chap. 36 #15. Thanks

Explanation / Answer

a.) is .0061000378 rad

b.) is .0122003027 rad

Use equation asin=m to findabove

c.) ?

d.) ?

e.) ?

f.) ?

Any Ideas for the rest? Any?

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