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**Hint:**Applying the formula of path difference in diffraction due to a single slit and using it with appropriate assumptions, we will be able to prove that intensity of diffraction fringes decreases as the increase in number. For the second proof we need to apply the formula angular width of the central maxima. Substituting the values of path difference for the first minima we will be able to prove the statement given.

**Formulae used:**Path difference in diffraction due to a single slit: $\Delta x = a\sin \theta $.

Where $a$ is the width of the slit and is expressed in meter $(m)$, $x$ is the path difference and is expressed in meter $(m)$ and $\theta $ diffraction angle and is expressed in degrees $(^\circ )$.

Angular width of central maxima: $\tan \theta = \dfrac{{{y_1}}}{D}$

Where ${y_1}$ is the angular width of the central maxima and is expressed in meter $(m)$ and $D$ is separation distance between slit and screen.

**Step by step solution:**

For the first proof we will use the formula for path difference which is $\Delta x = a\sin \theta $ .

We know that for maxima the value of $\Delta x$ is $(2x + 1)\dfrac{\lambda }{2}$.

Substituting this value in the equation of path difference we get,

$\Delta x = a\sin \theta $

$

\Delta x = a\sin \theta = (2x + 1)\dfrac{\lambda }{2} \\

\Rightarrow \Delta x = 3\dfrac{\lambda }{2} \\

$

Considering the slit being divided into three equal parts and considering the paths of the diffracted rays of the extreme ends we get,

$

\dfrac{2}{3}a\sin \theta = \dfrac{2}{3}a \times \dfrac{{3\lambda }}{{2a}} \\

\Rightarrow \Delta x = \lambda \\

$

From the above result we deduce that the path difference between the first two paths is $\dfrac{\lambda }{2}$. These two cancel out each other and therefore a minima is created.

The third point contributes to a point of intensity in between the two minima. This is called the first secondary minima.

Similarly it can be shown that the intensity of the secondary maxima decreases and they will go on decreasing as the number $(n)$ increases.

Therefore, the first statement “the intensity of diffraction fringes decreases as the other (n) increases” is proven.

The general maxima lies between the first minima which in turn on either side of the first minima.

For the first minima,

$a\sin \theta = \lambda $

But the diffraction angle is almost negligible. So we can assume that $\sin \theta \approx \theta $. Therefore we get $a\theta = \lambda $.

We know that the formula for the angular width of central maxima is $\tan \theta = \dfrac{{{y_1}}}{D}$. Here too $\tan \theta \approx \theta $.

Substituting the necessary values we get,

$

\tan \theta = \dfrac{{{y_1}}}{D} \\

\Rightarrow \theta = \dfrac{{{y_1}}}{D} \\

$

But,

$

a\theta = \lambda \\

\Rightarrow \theta = \dfrac{\lambda }{a} \\

$

Therefore,

$

\theta = \dfrac{{{y_1}}}{D} \\

\Rightarrow \dfrac{\lambda }{a}D = {y_1} \\

$

This is the total angular width of the secondary maxima on one side.

Since the maxima and minima are in symmetry, the value of angular width of the other side of the first minima $ = {y_2} = \dfrac{\lambda }{a}D = {y_1}$.

Therefore, the total width of the central maxima is $ = {y_2} + {y_1} = \dfrac{{2\lambda }}{a}D$.

Thus the first statement “angular width of the central maximum is twice that of the first order secondary maximum” is proven.

**Note:**The central maxima is due the constructive interference of the wavelets from all parts of the silt. The first secondary maxima is formed due to the interference of wavelets from the one third part of the slit. The second secondary maxima is formed due to the constructive interference of the wavelets from the one fifth part of the slit and so on.

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