In a double slit experiment two coherent beams have slightly different intensities I and $\delta I$. Show that the resultant intensity at the maxima is nearly 4I, while that at the minima is $(\delta I)^2/4I$

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, $(\frac{\delta I}{I})^{2}$

\[=4I+2\hspace{2mm}\delta I\]

\[\approx4I\]

For Intensity at Minima,

\[ I=I_1+I_2+2\sqrt{I_1 I_2}\cos\phi \]

\[ I_{min}=I+\left(I+\delta I\right)+2\sqrt{I(I+\delta I)}\cos 180\]

\[=2I+\delta I-2I\left(1+\frac{\delta I}{I}\right)^{\frac{1}{2}}\]

\[=2I+\delta I-2I\left[1+\frac{1}{2}\frac{\delta I}{I}+\frac{\frac{1}{2}.\frac{-1}{2}}{2}\left(\frac{\delta I}{I}\right)^2\right]\]

Ignoring the higher order terms, $(\frac{\delta I}{I})^{3}$

\[\approx\frac{{\left(\delta I\right)}^{2}}{4I}\]

How come we ignore $(\frac{\delta I}{I})^2$ in the maxima but not in the minima? The answer is relative size. Compared to 4I , $2\delta I$ is a very small term, so it was ignored. In the minima result we have only one term which is $(\delta I)^2$ even though it is very small it cannot be ignored.

When I have 100 rupees, one rupee can be ignored. When I do not have any money even 5 paise matters.

Binomial Theorem and approximation is a very important tool for many problems.