Question

A source S and a detector D are placed at a distance d apart. Big cardboard is placed at a distance \[\sqrt 2 d\] from the source and the detector as shown in the figure (below). The source emits a wave of wavelength = \[\dfrac{d}{2}\] which is received by the detector after reflection from the cardboard. It is found to be in phase with the direct wave received from the source. By what minimum distance should the cardboard be shifted away so that the reflected wave becomes out of phase with the direct wave?

Answer 1

Hint: The path difference can be defined as the distance travelled by the two waves at the meeting point. It can measure how much the wave is shifted from another.

Formula used:
Phase difference=\[2\pi \lambda \times \]path difference
Where \[\lambda \] is wavelength.

Complete step by step solution:
Given wavelength =\[\lambda = \dfrac{d}{2}\]
The distance between the source and the detector is d.
The distance of the cardboard from the source and detector is \[\sqrt 2 d\].

Image: Source and the detector shown in figure

The initial path difference is,
\[2\left( {\sqrt {{{\left( {\dfrac{d}{2}} \right)}^2} + 2{d^2}} } \right) - d\]
Let x be the distance to which it is shifted. Then the reflected wave becomes out of phase with the direct wave. Then the path difference is,
\[2\left( {\sqrt {{{\left( {\dfrac{d}{2}} \right)}^2} + {{\left( {2d + x} \right)}^2}} } \right) - d = \left( {2d + \dfrac{d}{4}} \right)\]

By solving, we get
\[\sqrt 2 d + x = 1.54d\]..............(by using\[\sqrt 2 = 1.414\])
\[\Rightarrow x = 1.54d - 1.414d\\ \therefore x{\rm{ = 0}}{\rm{.13d}}\]
Therefore, 0.13d is the minimum distance should the cardboard be shifted away so that the reflected wave becomes out of phase with the direct wave.

Note: The phase difference can be defined as the difference in the phase of the two travelling waves. The maximum path difference will be produced where the path difference between the two waves is equal to a whole number of the wavelengths.