Question

In a large room, a person receives direct sound waves from a source $120\,m$ away from him. He also receives waves from the $25\,m$ high ceiling at a point halfway between them. The two interfere constructively for wavelengths of:
A) $20\,m$, $\dfrac{{20}}{3}\,m$, $\dfrac{{20}}{5}\,m$
B) $10\,m$, $20\,m$, $30\,m$, etc.
C) $10\,m$, $5\,m$, $\dfrac{{10}}{3}\,m$, etc.
D) $15\,m$, $25\,m$, $35\,m$, etc.

Answer 1

Hint: The constructive interference of the wavelength of the sound waves can be determined by using the formula which shows the relation between the path difference and the phase difference. And by using this relation, the wavelength can be determined.

Formula used:
The phase difference of the sound waves can be given by,
$\delta = \dfrac{{2\pi }}{\lambda } \times \Delta x$
Where, $\delta $ is the phase difference of the sound waves, $\lambda $ is the wavelength of the sound waves and $\Delta x$ is the path difference.

Complete step by step solution:
Given that,
In a large room, a person receives direct sound waves from a source $120\,m$ away from him and also he receives waves from the $25\,m$ high ceiling at a point halfway between them.
Now, the path difference of the sound wave when the man receives the sound from two sides is given by the square root of the sum of the individual square of the distance and subtracted from the total distance, then
$\Delta x = 2\left( {\sqrt {{{25}^2} + {{60}^2}} } \right) - 120$
By squaring the terms inside the square root in the above equation, then
$\Delta x = 2\left( {\sqrt {625 + 3600} } \right) - 120$
By adding the terms inside the square root in the above equation, then
$\Delta x = 2\left( {\sqrt {4225} } \right) - 120$
By taking the square root in the above equation, then the above equation is written as,
$\Delta x = 2\left( {65} \right) - 120$
By multiplying the terms in the above equation, then the above equation is written as,
$\Delta x = 130 - 120$
By subtracting the terms in the above equation, then the above equation is written as,
$\Delta x = 10$
Now, the phase change is written as,
$\delta = 2n\pi $
By substituting the above equation in the phase difference formula, then
$2n\pi = \dfrac{{2\pi }}{\lambda } \times \Delta x$
By cancelling the same terms in the above equation, then the above equation is written as,
$n = \dfrac{{\Delta x}}{\lambda }$
By rearranging the terms in the above equation, then the above equation is written as,
$\lambda = \dfrac{{\Delta x}}{n}$

For, the constructive interference, the value of $n$ is $1$, $2$, $3$……… and also substituting the path difference in the above equation, then
$\lambda = 10$ for $n = 1$
For next interference,
$\lambda = 5$ for $n = 2$
For next interference,
$\lambda = \dfrac{{10}}{3}$ for $n = 3$

Hence, the option (C) is the correct answer.

Note: For the constructive interference the order of the interference is starting form $1$ and it is not starting from $0$. The order of the interference is continuously going with increasing order. If the order of the interference is increasing, the wavelength decreases.