In a double slit experiment two coherent beams have slightly different
intensities I and $\delta I$. Such that I >>$\delta I$. Show that the resultant intensity at the maxima is nearly 4I, while that
at the minima is $(\delta I)^2/4I$
Courtesy:ophysics.com
For Intensity at Maxima,
\[ I=I_1+I_2+2\sqrt{I_1 I_2}\cos\phi \]
\[I_{max}=I+\left(I+\delta I\right)+2\sqrt{I(I+\delta I)}\cos 0\]
\[=2I+\delta I+2I\left(1+\frac{\delta
I}{I}\right)^{\frac{1}{2}}\]
\[=2I+\delta I+2I\left[1+\frac{1}{2}\frac{\delta
I}{I}+\frac{\frac{1}{2}.\frac{-1}{2}}{2}\left(\frac{\delta
I}{I}\right)^2\right]\]
Using Binomial theorem, ignoring the higher order terms,
$(\frac{\delta I}{I})^{2}$
\[=4I+2\hspace{2mm}\delta I\]
\[\approx4I\]