91Ó°ÊÓ

$L=q\mathrm{λ}$

The waves are out of phase after the reflections

From the condition of constructive and destructive interference, find the smallest and second smallest value of q for which the two waves are out of phase.

Formulae are as follow:

1. $\mathrm{Ï•}=m\left(2\mathrm{Ï€}\right)$for constructive interference m = 0,1,2,..
2. $\mathrm{Ï•}=\left(2m+1\right)\mathrm{Ï€}$for destructive interference $m=0,1,23..$

The given condition states that the two waves are out of phase after the reflection. So, the interference should be destructive.

If the interference is destructive, then the path difference must be equal to .

So, from the figure, their path difference is , hence, from the condition, must be,

$L=\frac{\mathrm{λ}}{2}$…… (i)

But, it is given that,

$L=q\mathrm{λ}$

Substitute for into the equation (i)

$$\begin{gathered} q\mathrm{λ}=\frac{\mathrm{λ}}{2} \\ q=\frac{1}{2} \\ =0.5 \end{gathered}$$

Hence, the smallest value of q that put A and B exactly out of the phase with each other after the reflections is 0.5

For the second smallest value of,

$L=3\mathrm{λ}/2$…… (ii)

But,

$L=q\mathrm{λ}$

Substitute role="math" localid="1661349019513" $q\mathrm{λ}forL$into the equation (ii)

$$\begin{gathered} q\mathrm{λ}=3\mathrm{λ}/2 \\ q=1.5 \end{gathered}$$

Hence, the second smallest value of q that put and exactly out of the phase with each other after the reflections is 1.5 .