Consider an array of parallel wires with uniform spacing of 1.38 cm between cent
ID: 2036933 • Letter: C
Question
Consider an array of parallel wires with uniform spacing of 1.38 cm between centers. In air at 20.0°C, ultrasound with a frequency of 36.0 kHz from a distant source is incident perpendicular to the array. (Take the speed of sound to be 343 m/s.) (a) Find the number of directions on the other side of the array in which there is a maximum of intensity. (Enter an integer number of angles for which there is a maximum of intensity. An angle above the horizontal and an angle below the horizontal count as two separate angles. If there is a maximum at the horizontal, it also counts as an angle.) (b) Find the angle for each of these directions relative to the direction of the incident beam. (State the angles corresponding to maxima of intensity for each value of m. If there does not exist a maximum of intensity for a given value of m, enter 'NONE' in the answer blank.) (m Need Help? Read ItExplanation / Answer
a) this is diffraction grating
Wavelength is = 343m/s / (36*10^3)Hz = 9.53*10^-3 m
Maximum angle is 90 degree
No. of Directions ;-
m*wavelength = d * sin(theta)
m = ((1.38*10^-2)*sin(90)) / (9.53*10^-3)
m = ((1.38*10^-2)*(1)) / (9.53*10^-3)
m = 1.45
m = 1
Maximum no. of Directions on the other side of the array is,
= 2m + 1
= 2(1) + 1
= 3
b)
m = -2, the Angle is None
m = -1, the Angle is :-
theta = sin^-1(-1((9.53*10^-3)/(1.38*10^-2)))
theta = sin^-1(-0.6905797)
theta = -43.68 degree
m = 0, the angle is 0 degree
m = 1, the angle is 43.68 degree
m = 2, the angle is none