# Radio waves (419m) from a station travel along two paths from the transmitter to a house. The...

## Question:

Radio waves (419m) from a station travel along two paths from the transmitter to a house. The first path is direct, 27km, and the second path is by reflection from a mountain directly behind the house. No phase change occurs upon reflection from the mountain. What minimum distance from the mountain to the house cases destructive interference?

## Interference:

The term interference can be defined as the superposing of the two waves to form a larger, smaller, or same wave. The interference is categorized into types i.e., constructive and destructive interference.

## Answer and Explanation:

Given data:

• The radio waves travels the distance from the transmitter to house is {eq}\lambda = 419\,{\rm{m}} {/eq}
• The first path is {eq}d = 27\,{\rm{km}} {/eq}

The expression for the path difference between sound waves is

{eq}s = 2D\cdot\cdot\cdot{\rm(1)} {/eq}

The expression for the path difference in the destructive interference is

{eq}s = \left( {2n + 1} \right) \cdot \dfrac{\lambda }{2} {/eq}

• Here {eq}n = 0,1,2{\kern 1pt} .... {/eq} is the number of grits.

For the minimum condition, the number of grits is{eq}n = 0 {/eq}

Substituting the values in the above equation s,

{eq}s = \dfrac{\lambda }{2}\cdot\cdot\cdot{\rm(2)} {/eq}

Equating equation (1) and (2) as,

{eq}\begin{align*} 2D &= \dfrac{\lambda }{2}\\ D &= \dfrac{\lambda }{4}\\ D& = \dfrac{{\left( {419} \right)}}{4}\\ D &= 104.75\,{\rm{m}} \end{align*} {/eq}

Thus the minimum distance is {eq}D = 104.75\,{\rm{m}} {/eq}