Question

Two sources of coherent radio waves broadcasting in phase arel ocated as shown below. Each grid square is 0.5 m square, and the radio sources broadcast at λ=2.0m .

At Point A is the interference between the two sources constructive or destructive?

constructive destructive

At Point B is the interference between the two sources constructive or destructive?

constructive destructive

At Point C is the interference between the two sources constructive or destructive?

constructive destructive

At Point D is the interference between the two sources constructive or destructive?

constructive destructive

Answer 1

Concepts and reason

The concepts of coherence, path difference, constructive and destructive interference are required to solve the problem.

First, find the distance of the given point from each source. Then, calculate the difference between the two distances. Finally, use the condition of constructive and destructive interference and determine whether the interference is constructive or destructive.

Fundamentals

The path difference between the two interfering waves is given as,

$\Delta L = \left| {{L_1} - {L_2}} \right|$

Here, ${L_1}$is the distance of the point from one source and ${L_2}$ is the distance of the same point from the second source

The condition for interference is,

$\Delta L = n\lambda$

Here, $\lambda$ is the wavelength of the two sources.

For constructive interference, the value of m is the integral multiple of wavelength. That is,

$n = m$

Here, n is the order of pattern.

For destructive interference n is given as,

$n = m + 0.5$

(1)

Distance of the point A from the left source is,

${L_1} = 6a$

Here, a is the length of the one square grid.

Distance of the point A from the right source is,

${L_2} = 6a$

The path difference between the two interfering waves is given as,

$\Delta L = \left| {{L_1} - {L_2}} \right|$

Substitute 6a for ${L_1}$ and ${L_2}$ in the above equation $\Delta L = \left| {{L_1} - {L_2}} \right|$.

$\begin{array}{c}\\\Delta L = \left| {6a - 6a} \right|\\\\ = 0\\\end{array}$

The condition for interference is,

$\Delta L = n\lambda$

Replace m with n.

$\Delta L = m\lambda$

Rearrange the equation $\Delta L = m\lambda$ for m.

$m = \frac{{\Delta L}}{\lambda }$

Substitute 0 for $\Delta L$ in the above equation.

$\begin{array}{c}\\m = \frac{0}{\lambda }\\\\m = 0\\\end{array}$

The path difference is the integral multiple of the wavelength. Hence, the interference between the two waves at point A is constructive.

(2)

Distance of the point B from the left source is,

${L_1} = 3a$

Here, a is the length of the one square grid.

Distance of the point B from the right source is,

${L_2} = 9a$

The path difference between the two interfering waves is given as,

$\Delta L = \left| {{L_1} - {L_2}} \right|$

Substitute 3a for ${L_1}$ and 9a for ${L_2}$ in the above equation $\Delta L = \left| {{L_1} - {L_2}} \right|$.

$\begin{array}{c}\\\Delta L = \left| {3a - 9a} \right|\\\\ = 6a\\\end{array}$

Substitute 0.5 m for a in the above equation $\Delta L = 6a$.

$\begin{array}{c}\\\Delta L = 6\left( {0.5{\rm{ m}}} \right)\\\\ = 3.0{\rm{ m}}\\\end{array}$

The condition for interference is,

$\Delta L = n\lambda$

Rearrange the equation $\Delta L = n\lambda$ for n.

$n = \frac{{\Delta L}}{\lambda }$

Substitute 3.0 m for $\Delta L$ and 2.0 m for $\lambda$ in the above equation.

$\begin{array}{c}\\m = \frac{{3.0{\rm{ m}}}}{{2.0{\rm{ m}}}}\\\\ = 1.5\\\end{array}$

The above obtained value is matching with the term $n = m + 0.5$ and so, this interference can be considered as destructive interference.

If the path difference is equal to odd integral multiple of half of the wavelength then the interference is destructive. Therefore, the interference between the two waves at point B is destructive.

(3)

Distance of the point C from the left source is,

${L_1} = 7a$

Here, a is the length of the one square grid.

Distance of the point C from the right source is,

${L_2} = 5a$

The path difference between the two interfering waves is given as,

$\Delta L = \left| {{L_1} - {L_2}} \right|$

Substitute 7a for ${L_1}$ and 5a for ${L_2}$ in the above equation $\Delta L = \left| {{L_1} - {L_2}} \right|$.

$\begin{array}{c}\\\Delta L = \left| {7a - 5a} \right|\\\\ = 2a\\\end{array}$

Substitute 0.5 m for a in the above equation $\Delta L = 2a$.

$\begin{array}{c}\\\Delta L = 2\left( {0.5{\rm{ m}}} \right)\\\\ = 1.0{\rm{ m}}\\\end{array}$

Rearrange the equation $\Delta L = n\lambda$ for n.

$n = \frac{{\Delta L}}{\lambda }$

Substitute 1.0 m for $\Delta L$ and 2.0 m for $\lambda$ in the above equation.

$\begin{array}{c}\\m = \frac{{1.0{\rm{ m}}}}{{2.0{\rm{ m}}}}\\\\ = 0.5\\\end{array}$

The above obtained value is matching with the term $n = m + 0.5$ and so, this interference can be considered as destructive interference.

If the path difference is equal to odd integral multiple of half of the wavelength then the interference is destructive. Therefore, the interference between the two waves at point C is destructive.

(4)

Distance of the point D from the left source is,

${L_1} = 10a$

Here, a is the length of the one square grid.

Distance of the point C from the right source is,

${L_2} = 2a$

The path difference between the two interfering waves is given as,

$\Delta L = \left| {{L_1} - {L_2}} \right|$

Substitute 10a for ${L_1}$ and 2a for ${L_2}$ in the above equation $\Delta L = \left| {{L_1} - {L_2}} \right|$.

$\begin{array}{c}\\\Delta L = \left| {10a - 2a} \right|\\\\ = 8a\\\end{array}$

Substitute 0.5 m for a in the above equation $\Delta L = 8a$.

$\begin{array}{c}\\\Delta L = 8\left( {0.5{\rm{ m}}} \right)\\\\ = 4.0{\rm{ m}}\\\end{array}$

The condition for interference is,

$\Delta L = n\lambda$

Replace m with n.

$\Delta L = m\lambda$

Rearrange the equation $\Delta L = m\lambda$ for m.

$m = \frac{{\Delta L}}{\lambda }$

Substitute 4.0 m for $\Delta L$ and 2.0 m for $\lambda$ in the above equation.

$\begin{array}{c}\\m = \frac{{4.0{\rm{ m}}}}{{2.0{\rm{ m}}}}\\\\ = 2.0\\\end{array}$

This implies that the path difference is an integral multiple of wavelength. Therefore, the interference between the two waves at point D is constructive.

Ans: Part 1

At point A, the interference between the two waves is constructive.

Part 2

At point B, the interference between the two waves is destructive.

Part 3

At point C, the interference between the two waves is destructive.

Part 4

At point D, the interference between the two waves is constructive.