Question

Two harmonic waves of monochromatic light $Y_1 = a coswt$ and $y_2 = \text{a cos}\left( \text{wt}+\text{ }\!\!\theta\!\!\text{ } \right) $ ent some are superimposed on each other. Show that the maximum intensity in the interference pattern is four times the intensity due to each slit. Hence write the conditions for constructive and destructive interference in terms of the phase angle $ \text{ }\!\!\theta\!\!\text{ } $ .

Answer 1

Hint: A harmonic of such a wave is a wave with a frequency that is positive integer multiple of frequency of the original wave, known as the fundamental frequency. The intensity of light is directly proportional to the square of amplitude of the wave.
$ \text{I}=4{{\text{a}}^{2}}\text{ co}{{\text{s}}^{2}}\dfrac{\text{ }\!\!\theta\!\!\text{ }}{\text{2}} $
For constructive interference $ \text{ }\!\!\theta\!\!\text{ }=2\text{n }\!\!\pi\!\!\text{ } $
For destructive interference $ \text{ }\!\!\theta\!\!\text{ }=\left( 2\text{n}\pm 1 \right)\dfrac{\text{ }\!\!\pi\!\!\text{ }}{2} $.

Complete step by step solution
Two harmonic waves of monochromatic light
$ \begin{align}
  & {{\text{Y}}_{1}}=\text{a cos wt} \\
 & {{\text{Y}}_{2}}=\text{a cos}\left( \text{wt}+\text{ }\!\!\theta\!\!\text{ } \right) \\
\end{align} $
When they superimposed on each other
$ \begin{align}
  & \text{Y=}{{\text{Y}}_{1}}+{{\text{Y}}_{2}} \\
 & =\text{a cos wt}+\text{a cos}\left( \text{wt }+\text{ }\!\!\theta\!\!\text{ } \right) \\
\end{align} $
$ \begin{align}
  & =\text{a}\left[ \text{cos wt}+\cos \left( \text{wt }+\text{ }\!\!\theta\!\!\text{ } \right) \right] \\
 & =\text{2a cos}\dfrac{\text{ }\!\!\theta\!\!\text{ }}{\text{2}}\text{ cos}\dfrac{\text{w}}{2} \\
\end{align} $
The amplitude of the resultant displacement is $ 2\text{a cos}\dfrac{\text{ }\!\!\theta\!\!\text{ }}{\text{2}} $ . The intensity of light is directly proportional to the square of amplitude of the wave. The resultant intensity is given by:
 $ \text{I}=4{{\text{a}}^{2}}\text{ co}{{\text{s}}^{2}}\dfrac{\text{ }\!\!\theta\!\!\text{ }}{\text{2}} $
 $ \therefore $ Intensity $ =4\text{ I co}{{\text{s}}^{2}}\dfrac{\text{ }\!\!\theta\!\!\text{ }}{2} $
At the maximum, $ \text{ }\!\!\theta\!\!\text{ }=\pm 2\text{ n }\!\!\pi\!\!\text{ } $
 $ \therefore $ $ {{\cos }^{2}}\text{ }\dfrac{\text{ }\!\!\theta\!\!\text{ }}{2}=1 $
At the maxima, $ \text{I}=4\text{ }{{\text{I}}_{\text{o}}}=4\times $ intensity due to one slit
 $ \text{I}=\vartriangle {{\text{I}}_{\text{o}}}{{\cos }^{2}}\dfrac{\text{ }\!\!\theta\!\!\text{ }}{2} $
For constructive interference, I is maximum it is possible when $ {{\cos }^{2}}\left( \dfrac{\text{ }\!\!\theta\!\!\text{ }}{2} \right)=1 $ ;
 $ \dfrac{\text{ }\!\!\theta\!\!\text{ }}{2}=\text{n }\!\!\pi\!\!\text{ ; }\!\!\theta\!\!\text{ =2n }\!\!\pi\!\!\text{ } $
For destructive interference I is maximum i.e I=o
It is possible when
 $ \begin{align}
  & {{\cos }^{2}}\left( \dfrac{\text{ }\!\!\theta\!\!\text{ }}{2} \right)=0; \\
 & \dfrac{\text{ }\!\!\theta\!\!\text{ }}{2}=\dfrac{\left( 2\text{n}-1 \right)\text{ }\!\!\pi\!\!\text{ }}{2}, \\
 & \text{ }\!\!\theta\!\!\text{ }=\left( 2\text{n}\pm \right)\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \\
\end{align} $ .

Note
The frequency of the first harmonic is equal to wave speed divided by twice the length of the string. Its applications are clock, guitar, violin etc. For resonance in a taut string, the first harmonic is determined for a wave from one antinode and two nodes. Knowledge of nodes and antinodes is important in the case of harmonic waves.