The intensity at the central maximum of a double-slit interference pattern is $4I_1$. The intensity at the first minimum is zero. At what fraction of the distance from the central maximum to the first minimum is the intensity $I_1$? Assume an ideal double slit.

The average power transmission across one period of a wave, such as acoustic waves (sound) or electromagnetic waves like light or radio waves, is referred to as intensity. Intensity can be used to portray intensity in a variety of contexts.

The intensity of a perfect double slit can be calculated using the equation below.

$I_{\text{double}}=4I_1{\cos }^2\left(\frac{\pi d}{\lambda L}y\right)$

Let's substitute $I$text double with $I_1$obtain the ratio of the width between the centre high and the first minimum at which the intensity becomes $I_1$.

$\begin{array}{c}\frac{I_{\text{double}}}{4I_1}=\frac{I_1}{4I_1}={\cos }^2\left(\frac{\pi d}{\lambda L}y\right)\\ \end{array}$

${\cos }^2\left(\frac{\pi d}{\lambda L}y\right)=\frac{1}{4}$

The original number of both sides yields

$\begin{array}{c}\cos \left(\frac{\pi d}{\lambda L}y\right)=\frac{1}{2}\\ {\cos }^{-1}\left(\frac{1}{2}\right)=\frac{\pi d}{\lambda L}y\end{array}$

to separate y, change the equation

$y=\frac{\lambda L}{\pi d}{\cos }^{-1}\left(\frac{1}{2}\right)$

This is the location where the intensity becomes $I_1$, However, since our purpose is to find the ratio of this location to the first minimum position, we'll apply the phrase below to find the first minimum position.

$y_m^{\mathrm{\prime }}=\left(m+\frac{1}{2}\right)\frac{\lambda L}{d}m=0,1,2,\dots$

As a result, that the very first minimum's position is

$y_0^{\mathrm{\prime }}=\frac{\lambda L}{2d}$

As a result, the width fraction between the centre best and the first minimum when the intensities is increased. $I_1$is

$$\begin{gathered} \frac{y}{y_0^{\mathrm{\prime }}}=\frac{\frac{\lambda L}{\pi d}{\cos }^{-1}\left(0.5\right)}{\frac{\lambda L}{2d}} \\ =\frac{2}{3}=0.666 \end{gathered}$$