Question

Every day at the end of recess, Principal Umbridge stands in a 1 meter wide doorway and blows a whistle of frequency 171.5 hz. Standing on the fence 10 m away and perpendicular to the doorway is a mischievous boy. He decides to stay on the playground, pretending not to hear. When the principal comes to get him, he explains to her that he must have been at a minima in the single-slit interference pattern caused by the sound wave diffracting through the doorway. The principal says that she will accept his explanation if he can tell her the angle of the first order minima and the distance along the fence that he was at relative to the center of the doorway.

1. What is the wavelength of the whistle?

2. What angle should the boy tell the principal? (Hint: do not use small angle approximation)

3. What distance along the fence should he tell her? (Hint: do not use small angle approximation)

Answer 1

We knwo that the condition for minima in single slit diffraction is  

d sin theta = m*lambda

where d is slit width = wide of the door = 1 m ,

lambda is wavelength of sound wave (wistle)

m = 1

given frequency of the wistle is f = 171.5 Hz and  

distance from the principal to the student is L = 10 m

1.

we know that v = lambda*f

lambda = v/f

lambda = 343 /171.5 m = 2 m

2.

substituting the values for theta value sin theta = m*lambda/d

theta = arc sin (m*lambda/d)

theta = arc sin(1*2/1)

theta = arc sin (2)

which is not possible so the student is at the maximum position that is  

d sin theta = (m+0.5)*lambda

theta = arc sin (0.5*lambda/d)

theta = arc sin(0.5*2/1)

theta = arc sin (1) = 90 degrees

3.

the distance is sin theta = y/l ==> y = l*sin 90 = 10 m