A single slit with a width of 0.100 mm is illuminated by lightwitha wavelength 610 nm.
(a) What is the angular location of the firstdiffractionminimum?
? rad
(b) What is the angular location of the seconddiffractionminimum?
? rad
(c) Diffraction maxima occur approximately halfway betweenadjacentdiffraction minima. Using your results from (a) and (b),calculatethe approximate angular location of the first diffractionpeak tothe left or right of the central diffraction peak.
? rad
(d) To obtain the "exact" locations of diffraction maxima, valuesofalpha are required at which intensity maxima occur for asingleslit. An equation for these values of alpha are obtainedbydifferentiating Eq. 36-5 with respect to alpha and equatingtheresult to zero. Two equations are obtained: (1) The firstequation,sin / = 0, is the condition that must be satisfied whenthere is adiffraction minima. (2) The second equation of the form(sin )n1(cos )n2 = ()n3 is the condition that must be satisfiedwhen thereis a diffraction maxima. What are the values of theintegers n1, n2and n3, at least one of which is negative?
n1= ?
n2= ?
n3= ?
(e) It is not possible to exactly solve the equation fordiffractionmaxima to obtain a general formula for the location ofalldiffraction maxima; however, the values of alpha that satisfythisequation can be found graphically by plotting the curve y =(sin )n1(cos )n2 and the curve y = ()n3 and determining where thetwo curvesintersect. The smallest two values of alpha that satisfytheequation are alpha = 0 and alpha = 4.4934. The first solutionisexact, and the second and all others are only approximate. Usingthesolution alpha = 4.4934, solve for the angular location thetaof thediffraction peak adjacent to the central peak.
? rad
(f) Calculate the percent error between the actual value ofthetacalculated in (e) and the approximate value calculated in(c).
????%
I have a and b need help with the rest. This is a modifiedversionof Chap. 36 #15. Thanks
A single slit with a width of 0.100 mm is illuminated by lightwitha wavelength 610 nm.
(a) What is the angular location of the firstdiffractionminimum?
? rad
(b) What is the angular location of the seconddiffractionminimum?
? rad
(c) Diffraction maxima occur approximately halfway betweenadjacentdiffraction minima. Using your results from (a) and (b),calculatethe approximate angular location of the first diffractionpeak tothe left or right of the central diffraction peak.
? rad
(d) To obtain the "exact" locations of diffraction maxima, valuesofalpha are required at which intensity maxima occur for asingleslit. An equation for these values of alpha are obtainedbydifferentiating Eq. 36-5 with respect to alpha and equatingtheresult to zero. Two equations are obtained: (1) The firstequation,sin / = 0, is the condition that must be satisfied whenthere is adiffraction minima. (2) The second equation of the form(sin )n1(cos )n2 = ()n3 is the condition that must be satisfiedwhen thereis a diffraction maxima. What are the values of theintegers n1, n2and n3, at least one of which is negative?
n1= ?
n2= ?
n3= ?
(e) It is not possible to exactly solve the equation fordiffractionmaxima to obtain a general formula for the location ofalldiffraction maxima; however, the values of alpha that satisfythisequation can be found graphically by plotting the curve y =(sin )n1(cos )n2 and the curve y = ()n3 and determining where thetwo curvesintersect. The smallest two values of alpha that satisfytheequation are alpha = 0 and alpha = 4.4934. The first solutionisexact, and the second and all others are only approximate. Usingthesolution alpha = 4.4934, solve for the angular location thetaof thediffraction peak adjacent to the central peak.
? rad
(f) Calculate the percent error between the actual value ofthetacalculated in (e) and the approximate value calculated in(c).
????%
I have a and b need help with the rest. This is a modifiedversionof Chap. 36 #15. Thanks
