Question

The figure shows the circular wave fronts emitted by two sources.

A) Give the distances from points P, Q and R to the point 1 in indicated order as multiples of lambda.

B) Give the distances from points P, Q and R to the point 2 in indicated order as multiples of lambda.

C) Indicate whether the interference at the point P is constructive or destructive.

D) Indicate whether the interference at the point Q is constructive or destructive.

E) Indicate whether the interference at the point R is constructive or destructive.

Answer 1

Concepts and reason

The concept required to solve the given problem is interference of sound waves.

Initially calculate the distances from points P, Q, R to the point 1 and 2 in terms of wavelength. Later on, find out whether the interference at the points P and Q is constructive or destructive. Finally, find out whether the interference at point R is constructive or destructive.

Fundamentals

Wave interference- It is the phenomenon that occurs when two waves superimposed while traveling along the same medium.

Constructive interference: If the amplitude of sound waves is superimposed in such a manner that their amplitudes is add up, so the interference is known as constructive interference.

Destructive interference: If the amplitude of sound waves is superimposed in such a manner that their amplitudes are out of phase, so the interference is known as constructive interference.

(A)

The closest distance between two circular wave fronts is given by,

$\beta = \frac{{\lambda D}}{d}$ …… (1)

Here, $\beta$ is the fringe width, $\lambda$ is the wavelength of sound wave, d is the distance between two slits, and D is the separation between slit and screen.

As the separation between two circular wave fronts is proportional to wavelength, so the distance of point P from the source 1 is,

${r_{1P}} = 3\lambda$

The distance of point Q from point 1 is,

${r_{1Q}} = 3.5\lambda$

The distance of point R from point 1 is,

${r_{1R}} = 2.5\lambda$

(B)

The closest distance between two circular wave fronts is given by,

$\beta = \frac{{\lambda D}}{d}$ …… (1)

Here, $\beta$ is the fringe width, $\lambda$ is the wavelength of sound wave, d is the distance between two slits, and D is the separation between slit and screen.

As the separation between two circular wave fronts is proportional to wavelength, so the distance of point P from the source 2 is,

${r_{2P}} = 4\lambda$

The distance of point Q from the source 2 is,

${r_2}_Q = 2\lambda$

Similarly, the distance of point R from the source 2 is,

${r_{2R}} = 3.5\lambda$

(C)

If the path difference between the wave fronts coming from two points 1 and 2 is equal to integer multiple of wavelength, then the two waves are said to be constructively interference and if it is equal to the odd multiple of wavelength, then the waves are said to be destructively interference.

The path difference between the two waves reaching at point P is,

${\rm{Path difference = }}{r_{{\rm{2p}}}} - {r_{1p}}$

Here, ${r_{{\rm{2p}}}}$ is the distance of point P from source 2, and ${r_{1p}}$ is the distance of point P from source 1.

Substitute $4\lambda$ for ${r_{2p}}$ and $3\lambda$ for ${r_{1p}}$

$\begin{array}{c}\\{\rm{Path difference = }}\left| {{r_{{\rm{2p}}}} - {r_{1p}}} \right|\\\\ = 4\lambda - 3\lambda \\\\ = \left( 1 \right)\lambda \\\end{array}$

As the path difference is integer multiple of wavelength, thus the interference is constructive.

(D)

The path difference between the two waves reaching at point Q is,

${\rm{Path difference = }}{r_{{\rm{2Q}}}} - {r_{1Q}}$

Here, ${r_{{\rm{2Q}}}}$ is the distance of point Q from source 2, and ${r_{1Q}}$ is the distance of point Q from source 1.

Substitute $3.5\lambda$ for ${r_{2Q}}$ and $2\lambda$ for ${r_{1Q}}$

$\begin{array}{c}\\{\rm{Path difference = }}\left| {{r_{{\rm{2Q}}}} - {r_{1Q}}} \right|\\\\ = \left| {3.5\lambda - 2\lambda } \right|\\\\ = \left( {1.5} \right)\lambda \\\end{array}$

As the path difference is equal to half integer multiple of wavelength, thus the interference is destructive.

(E)

The path difference between the two waves reaching at point R is,

${\rm{Path difference = }}{r_{{\rm{2R}}}} - {r_{1R}}$

Here, ${r_{{\rm{2R}}}}$ is the distance of point R from source 2, and ${r_{1R}}$ is the distance of point R from source 1.

Substitute $2.5\lambda$ for ${r_{2R}}$ and $3.5\lambda$ for ${r_{1R}}$

$\begin{array}{c}\\{\rm{Path difference = }}\left| {{r_{{\rm{2R}}}} - {r_{1R}}} \right|\\\\ = \left| {2.5\lambda - 3.5\lambda } \right|\\\\ = \left( 1 \right)\lambda \\\end{array}$

As the path difference is equal to integer multiple of wavelength, thus the interference is constructive.

Ans: Part A

The distances from point P, Q and R to the point 1 is $3\lambda ,{\rm{ 3}}{\rm{.5}}\lambda {\rm{, and 2}}{\rm{.5}}\lambda$ .

Part B

The distance of points P, Q and R from point 2 are $4\lambda ,{\rm{ 2}}\lambda {\rm{, 3}}{\rm{.5}}\lambda$ .

Part C

The interference at the point P is constructive.

Part D

The interference at the point Q is destructive.

Part E

The interference at the point R is constructive.