Q.168. Interpherence, Young's Double-slit Formula

Question 168.
A police cruiser has an unusual radar speed trap set up. It has two transmitting antennae at the edge of a main road that runs north and south. One antenna is 2.0 m[west] of the other, and they are both fed from a common transmitter with a frequency of 3.0 x10^9 Hz. These antennae can be considered as point sources of continuous radio waves. The trap is set for cars travelling south. A student drives his car along an east-west road that crosses the main road at a level intersection 100 m[north] of the radar trap. The student's car has a "radar detector", so he hears a series of beeps as he drives west through the intersection. If the time interval between successive quiet spots is 1/5 s, as he crosses the main road, what is his speed?

Answer 168.
This is the problem of Young's double-slit interference experiment.

d = 2.0 m
D = 100 m
λ = c/f = 3x10^8 / 3 x 10^9 = 10^(-1) m
distance between successive destructive interference,
x = λD/d = 10^(-1) * 100/2 = 5 m
=> speed of the car
= 5/(1/5) m/s
= 25 m/s
= 25 * 3.6 km/h
= 90 km/h.

Link to YA!