Interference of Waves — Question 4

Question 4

A slower inserted segment can reverse an interference maximum

Two coherent paths initially arrive in phase at frequency $f=20\,\mathrm{Hz}$. In one path, a length $\ell$ with speed $30\,\mathrm{m/s}$ is replaced by a region with speed $20\,\mathrm{m/s}$. Assume amplitudes stay equal, reflection is suppressed, and only the travel-time phase changes.

1. Find the extra delay and phase lag.

2. Find the smallest positive $\ell$ that converts the maximum into a null.

Question 4 - Solution

A slower inserted segment can reverse an interference maximum

Method and mathematical working

Calculate the additional travel time caused by replacement of the segment. The first cancellation requires a half-cycle of extra delay.$$\begin{aligned}\Delta t&=\ell\left(\frac1{20}-\frac1{30}\right)=\frac{\ell}{60},\\2\pi f\Delta t&=\pi\Rightarrow\ell=\frac{60}{2f}.\\[4pt]\Delta t&=\frac1{2(20)}=0.025\,\mathrm s,\qquad\ell=60(0.025)=1.50\,\mathrm m.\end{aligned}$$

(a) Calculation and conclusion

The extra delay is $\Delta t=\ell(1/20-1/30)=\ell/60$. The added phase lag magnitude is $2\pi f\Delta t=2\pi\ell/3$, with $\ell$ in meters.

(b) Interpretation and limits

A null requires a phase difference $\pi$ modulo $2\pi$. The smallest length is $\ell=1.50\,\mathrm{m}$, corresponding to delay $.025\,\mathrm{s}$, one half-cycle. This conclusion relies on the explicitly maintained equal amplitudes; losses or unmatched interfaces could prevent a complete null.

Physical check

The delay is positive because the inserted region is slower. Further nulls occur for odd multiples of this first length, assuming unchanged amplitudes and negligible interface effects.