JEE Main & Advanced Physics Wave Optics / तरंग प्रकाशिकी Resultant Amplitude and Intensity

Let us consider two waves that have the same frequency but have a certain fixed (constant) phase difference between them. Their super position shown below

Let the two waves are

\[{{y}_{1}}={{a}_{1}}\sin \,\omega \,t\] and \[{{y}_{2}}={{a}_{2}}\sin \,(\omega \,t+\varphi )\]

where \[{{a}_{1}},\,{{a}_{2}}=\] Individual amplitudes,

\[\phi =\] Phase difference between the waves at an instant when they are meeting a point.

(1) Resultant amplitude : The resultant wave can be written as \[y=A\,\,\sin \,(\omega t+\theta )\]

where A = resultant amplitude \[=\sqrt{a_{1}^{2}+a_{2}^{2}+2{{a}_{1}}{{a}_{2}}\cos \varphi }\]

(2) Resultant intensity : As we know intensity \[\propto {{(\text{Amplitude})}^{2}}\]

\[\Rightarrow \] \[{{I}_{1}}=ka_{1}^{2},\,{{I}_{2}}=ka_{2}^{2}\] and \[I=k{{A}^{2}}\] (k is a proportionality constant). Resultant intensity  \[I={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \varphi \]

For two identical source \[{{I}_{1}}={{I}_{2}}={{I}_{0}}\]\[\Rightarrow \]\[I={{I}_{0}}+{{I}_{0}}+2\sqrt{{{I}_{0}}{{I}_{0}}}\cos \varphi \]

\[=4{{I}_{0}}{{\cos }^{2}}\frac{\varphi }{2}\] \[[1+\cos \theta =2{{\cos }^{2}}\frac{\theta }{2}]\]