91Ó°ÊÓ

The intensity I on the screen is;

$I\left(\mathrm{γ}\right)=I_0{\left(\frac{\sin {\displaystyle }\mathrm{γ}}{\mathrm{γ}}\right)}^2$

Here, $\mathrm{γ}=\mathrm{β}/2$

$\mathrm{β}$is the phase difference between two waves received from the slit's two endpoints, and $l_0$is the intensity in a straight-ahead direction.

$\frac{dl}{d{\mathrm{γ}}_{{\mathrm{γ}}_0}}=0$

Now the derivative of I with respect to $\mathrm{γ}$

$\begin{array}{r}\frac{dl}{d\mathrm{γ}}=2I_0\frac{\sin \mathrm{γ}}{\mathrm{γ}}â‹\dots \frac{\mathrm{γ}\cos \mathrm{γ}−\sin \mathrm{γ}}{{\mathrm{γ}}^2}\\ =2I_0\frac{\sin \mathrm{γ}(\mathrm{γ}\cos \mathrm{γ}−\sin \mathrm{γ})}{{\mathrm{γ}}^3}\end{array}$

As a result, the following equations for the local extremum are obtained:

1. $\sin \mathrm{γ}=0$
2. $\mathrm{γ}\cos \mathrm{γ}-\sin \mathrm{γ}=0$

Therefore;

$$\begin{gathered} \sin \left(\frac{\mathrm{β}}{2}\right)=0 \\ \frac{\mathrm{β}}{2}=\frac{\mathrm{π}}{\mathrm{λ}}a\sin \mathrm{θ}=m\mathrm{π} \end{gathered}$$

These angles correspond to destructive interference points, i.e., minima for which I=0. As a result, the maxima will follow the second equation;

$\tan \mathrm{γ}=\mathrm{γ}$

$f\left({\mathrm{γ}}_0\right)={\mathrm{γ}}_0$is a point called a fixed point of the function f.

Now approximating the solution, at a point ${\mathrm{γ}}_1$and is in the vicinity of ${\mathrm{γ}}_0$

So, the sequence ${\mathrm{γ}}_n$is;

$$\begin{gathered} {\mathrm{γ}}_2=f\left({\mathrm{γ}}_1\right) \\ ... \\ {\mathrm{γ}}_n=f\left({\mathrm{γ}}_{n-1}\right) \end{gathered}$$

As (f) is continuous as ${\mathrm{γ}}_1$is close to ${\mathrm{γ}}_0$

$\begin{array}{r}\underset{n→\mathrm{∞}}{lim} {\mathrm{γ}}_n={\mathrm{γ}}_0\\ \left|f^{\mathrm{′}}\left({\mathrm{γ}}_0\right)\right|<1\end{array}$

So, the sequence is;

$$\begin{gathered} {\mathrm{γ}}_2=4.463 \\ {\mathrm{γ}}_3=4.492 \\ {\mathrm{γ}}_4=4.493 \\ ... \\ {\mathrm{γ}}_1=\underset{n→\mathrm{∞}}{lim} {\mathrm{γ}}_n \\ =4.4934 \end{gathered}$$

$\mathrm{Ï€},2\mathrm{Ï€},3\mathrm{Ï€}$

And the midpoints of the three points are;

$\begin{array}{r}\frac{3\mathrm{Ï€}}{2}=4.712≠{\mathrm{γ}}_1\\ \frac{5\mathrm{Ï€}}{2}=7.854≠{\mathrm{γ}}_{′â\P IJ}\end{array}$

$$\begin{gathered} \mathrm{γ}=\frac{\mathrm{β}}{2}=\mathrm{π}\frac{a}{\mathrm{λ}}\sin \mathrm{θ}=12\mathrm{π}\sin \mathrm{θ} \\ As\mathrm{γ}>0 \\ \mathrm{θ}arc\sin \left(\frac{\mathrm{γ}}{12\mathrm{π}}\right) \end{gathered}$$

As;

The first minimum $\mathrm{γ}=\mathrm{π}$

The first maximum beyond central maximum $\mathrm{γ}=4.4934$

The second minimum $\mathrm{γ}=2\mathrm{π}$

$$\begin{gathered} {\mathrm{θ}}_{1m}=\arcsin \left(\frac{\mathrm{π}}{12\mathrm{π}}\right)={4.8}^{∘} \\ {\mathrm{θ}}_{1M}=\arcsin \left(\frac{4.4934}{12\mathrm{π}}\right)={6.8}^{∘} \\ {\mathrm{θ}}_{2m}=\arcsin \left(\frac{2\mathrm{π}}{12\mathrm{π}}\right)={9.6}^{∘} \end{gathered}$$