Question

(a) Show that the values of a at which intensity maxima for single-slit diffraction occur can be found exactly by differentiating Eq. 36-5 with respect to a and equating the result to zero, obtaining the condition $\mathit{t}\mathit{a}\mathit{n}\mathit{\alpha }\mathbf{=}\mathbf{\alpha }$. To find values of a satisfying this relation, plot the curve $\mathbf{y}\mathbf{=}\mathbf{tan\alpha }$ and the straight line $\mathbf{y}\mathbf{=}\mathbf{\alpha }$ and then find their intersections, or use $\mathbf{\alpha }$calculator to find an appropriate value of a by trial and error. Next, from $\mathbf{\alpha }\mathbf{=}\mathbf{(}\mathbf{m}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{)}\mathbf{\pi }$, determine the values of $\mathit{m}$ associated with the maxima in the singleslit pattern. (These m values are not integers because secondary maxima do not lie exactly halfway between minima.) What are the (b) smallest and (c) associated , (d) the second smallest $\mathbf{\alpha }$(e) and associated , (f) and the third smallest (g) and associated ?

Short answer

1. The value is$\tan \alpha =\alpha$.
2. The value is$y=\tan \alpha$
3. The value is$\left(m=-\frac{1}{2}\right)$.
4. The value is $\alpha =4.493rad$.
5. The value is$\left(m\right)=0.93$.
6. The value is $\left(a\right)=7.725rad$.
7. The value is $\left(m\right)=1.96$.

Step-by-step solution

Introduction

As the intensity increases, the diffraction maximum becomes narrower as well as more intense. When you have 600 slits, the maxima are very sharp and bright and permit high-resolution separation of the maxima for different wavelengths. Such a multiple-slit is called a diffraction grating.

Concept

In single slit diffraction,

Intensity is given as $I=I_m\frac{{\text{sin}}^2\alpha }{{\alpha }^2}$

(a) Determine the value of intensity maxima

Differentiating the above with respect to a ,

$\frac{dI}{d\alpha }=2I_m\frac{\sin \alpha }{{\alpha }^3}\left(\alpha \cos \alpha -\text{sin}\alpha \right)$

For maxima and minima$\frac{dI}{d\alpha }=0$

So either

$\alpha =0$or $\text{sin}\alpha =0$or $\alpha \cos \alpha -\text{sin}\alpha =0$

But $\alpha =0$so $\text{sin}\alpha =0$

So $\alpha =m\pi$

Again, if $\alpha \cos \alpha -\text{sin}\alpha =0$

Or $\tan \alpha =\alpha$

Since ${\left(\frac{d^{2}I}{d{\alpha }^2}\right)}_{\alpha =\tan \alpha }=negative$

There is a maxima at $\tan \alpha =\alpha$.

(b) Determine the value of smallest

Let $y=\tan \alpha$ where $y=\alpha$

From the graph,

The smallest value of $\alpha =0$.

Hence, the value is$\alpha =0$.

(c) Determine the value of associated

As,$\tan \alpha =\alpha$,

localid="1664272360306" $\alpha =\left(m+\frac{1}{2}\right)\pi$

For central maximum, $\alpha =0$

Hence, the value is $\left(m=-\frac{1}{2}\right)$

(d) Determine the value of second smallest α

The second smallest $\alpha =4.493rad$

Hence, the value is$\alpha =4.493rad$.

(e) Determine the value of associated m

Associated $\left(m\right)=\frac{a}{\pi }-\frac{1}{2}$

$\left(m\right)=0.93$

Hence, the value is $\left(m\right)=0.93$

(f) Determine the value of third smallest α

The third smallest $\left(a\right)=7.725rad$

Hence, the value is $\left(a\right)=7.725rad$

(g) Determine the value of associated 

Associated $\left(m\right)=\frac{a}{\pi }-\frac{1}{2}$

$\left(m\right)=1.96$

Hence, the value is $\left(m\right)=1.96$