The intensity of light in the Fraunhofer diffraction pattern of a single slit is given by Eq. (36.5). Let$\mathit{\gamma }\mathbf{=}\mathit{\beta }\mathbf{/}\mathbf{2}$. (a) Show that the equation for the values of role="math" localid="1668221820264" $\mathit{\gamma }$at which I is a maximum is tan $\mathit{\gamma }\mathbf{=}\mathit{\gamma }$. (b) Determine the two smallest positive values of $\mathit{\gamma }$that are solutions of this equation. (Hint: You can use a trial-and-error procedure. Guess a value of $\mathit{\gamma }$and adjust your guess to bring tan $\mathit{\gamma }$closer to$\mathit{\gamma }$. A graphical solution of the equation is very helpful in locating the solutions approximately, to get good initial guesses.) (c) What are the positive values of $\mathit{\gamma }$for the first, second, and third minima on one side of the central maximum? Are the $\mathit{\gamma }$values in part (b) precisely halfway between the $\mathit{\gamma }$values for adjacent minima? (d) If a = 12$\mathit{\lambda }$, what are the angles$\mathit{\theta }$(in degrees) that locate the first minimum, the first maximum beyond the central maximum, and the second minimum?

The intensity I on the screen is;

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

Here, $\gamma =\beta /2$

$\beta$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.

$\frac{dl}{d{\gamma }_{{\gamma }_0}}=0$

Now the derivative of I with respect to $\gamma$

$\begin{array}{r}\frac{dl}{d\gamma }=2I_0\frac{\sin \gamma }{\gamma }\cdot \frac{\gamma \cos \gamma -\sin \gamma }{{\gamma }^2}\\ =2I_0\frac{\sin \gamma (\gamma \cos \gamma -\sin \gamma )}{{\gamma }^3}\end{array}$

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

1. $\sin \gamma =0$
2. $\gamma \cos \gamma -\sin \gamma =0$

Therefore;

$$\begin{gathered} \sin \left(\frac{\beta }{2}\right)=0 \\ \frac{\beta }{2}=\frac{\mathrm{\pi }}{\lambda }a\sin \theta =m\mathrm{\pi } \end{gathered}$$

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

$\tan \gamma =\gamma$

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

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

So, the sequence ${\gamma }_n$is;

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

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

$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim} {\gamma }_n={\gamma }_0\\ \left|f^{\mathrm{\prime }}\left({\gamma }_0\right)\right|<1\end{array}$

So, the sequence is;

$$\begin{gathered} {\gamma }_2=4.463 \\ {\gamma }_3=4.492 \\ {\gamma }_4=4.493 \\ ... \\ {\gamma }_1=\underset{n\to \mathrm{\infty }}{lim} {\gamma }_n \\ =4.4934 \end{gathered}$$

$\mathrm{\pi },2\mathrm{\pi },3\mathrm{\pi }$

And the midpoints of the three points are;

$\begin{array}{r}\frac{3\pi }{2}=4.712\ne {\gamma }_1\\ \frac{5\pi }{2}=7.854\ne {\gamma }_{\prime \prime }\end{array}$

$$\begin{gathered} \gamma =\frac{\beta }{2}=\pi \frac{a}{\lambda }\sin \theta =12\pi \sin \theta \\ As\gamma >0 \\ \theta arc\sin \left(\frac{\gamma }{12\mathrm{\pi }}\right) \end{gathered}$$

As;

The first minimum $\gamma =\mathrm{\pi }$

The first maximum beyond central maximum $\gamma =4.4934$

The second minimum $\gamma =2\mathrm{\pi }$

$$\begin{gathered} {\theta }_{1m}=\arcsin \left(\frac{\pi }{12\pi }\right)={4.8}^{\circ } \\ {\theta }_{1M}=\arcsin \left(\frac{4.4934}{12\pi }\right)={6.8}^{\circ } \\ {\theta }_{2m}=\arcsin \left(\frac{2\pi }{12\pi }\right)={9.6}^{\circ } \end{gathered}$$