Question

he 22 In Fig. 17-38, sound with a s 40.0 cm wavelength travels right ve ward from a source and through a x.1 tube that consists of a straight por- tion and a half-circle. Part of the sound wave travels through the half- circle and then rejoins the rest of the wave, which goes directly through the straight portion. This rejoining results in interference. What is the smallest radius r that results in an in- tensity minimum at the detector?

Answer 1

path difference between 2 interfering waves=pir-2r=1.14r

For intensity minima destructive interference must take place

Path difference=wavelength/2

1.14r= 40/2

r=17.5cm