### Diffraction due to a plane diffraction grating or N- Parallel slits

A diffraction grating (or $N$-slits) consists of a large number of parallel slits of equal width and separated from each other by equal opaque spaces.

It may be constructed by ruling a large number of parallel and equidistance lines on a plane glass plate with the help of a diamond point. the duplicates of the original grating are prepared by pouring a thin layer of colloidal solution over it and then allowed to Harden. This layer is then removed from the original grating and fixed between two glass plates which serve as a plane transmission grating. Generally, A plane transmission grating has 10000 to 15000 lines per inch.
 Diffraction due to N- slits OR Grating

Theory→

Since plane diffraction grating is an $N$-slit arrangement, the deflection pattern due to it will be the combined diffraction effect of all such slits. Let a plane wavefront of monochromatic light be incident normally on the $N$-parallel slit of the gratings. Each point within the slits then sends out secondary wavelets in all directions.

Let $e$ be the width of each slipped and $d$ be the separation between any two consecutive slits then $(e+d)$ is known as the grating element. The diffracted ray from each slit, then $(e+d)$ is knowns as the grating element. The diffracted ray from each slit is focussed at a point $P$ on the screen $XY$ with the help of a convex lens $L$.

Expression for Intensity→

Let $S_{1}, S_{2}, S_{3},.......$ be the middle point of each slit and $S_{1}M_{1}, S_{2}M_{2}, S_{3}M_{3}, ........S_{N-1}M_{N-1}$ be the perpendicular drawn as shown in the figure.The waves diffracted from each slit are equivalent to a single wave amplitude:

$R=\frac{A\:sin\alpha}{\alpha} \qquad(1)$

The path difference between the waves from slit $S_{1}$ and $S_{2}$ is

$S_{2}M_{1}=(e+d)sin\theta$

The path difference between the waves from slit $S_{2}$ and $S_{3}$ is

$S_{3}M_{2}=(e+d)sin\theta$

The path difference between the waves from slit $S_{n-1}$ and $S_{n}$ is

$S_{N}M_{N-1}=(e+d)sin\theta$

Thus, it is obvious that the path difference between all the consecutive waves is the same and equal to $(e+d)sin\theta$

The corresponding phase difference

$\Delta \phi=\frac{2\pi}{\lambda}(e+d)sin\theta \qquad(2)$

Let $\Delta \phi=2\beta$

$\beta=\frac{\pi}{\lambda}(e+d)sin\theta \qquad(3)$

Thus, the resultant amplitude at $P$ is the resultant amplitude of $N$ waves, each of amplitude $R$ and its common phase difference is $(2\beta)$

$R'=\frac{R \: sin \left( \frac{2N\beta}{2} \right) }{sin \left( \frac{2\beta}{2} \right)} \qquad(4)$

The resultant amplitude at $P$

$R'=\frac{R \: sin N\beta}{sin \beta}$

Where $R=\frac{A\: sin\alpha}{\alpha}$. Now substitute the value of $R$ in the above equation and we get

$R'= \frac{A\: sin\alpha}{\alpha} \frac{ \: sin N\beta}{sin \beta} \qquad(5)$

The resultant intensity at $P$

$I=R'^{2}$

$I=\frac{A^{2} \: sin^{2} \alpha}{\alpha^{2}} \frac{ \: sin^{2} N\beta}{sin^{2} \beta} \qquad(6)$

The factor $\frac{A^{2} \: sin^{2} \alpha}{\alpha^{2}}$ gives the intensity pattern due to diffraction from a single slit while the factor $\frac{ \: sin^{2} N\beta}{sin^{2} \beta}$ gives the distribution of intensity due to interference from all the $N$-slit

Principle Maxima→

The intensity will be maximum when $sin\beta=0$ or $\beta=\pm n\pi$

Where $n=0,1,2,3,.....$

But under this condition, $sinN\beta$ is also equal to zero. Hence term $\frac{sin N\beta}{sin \beta}$ can be solve by

$\lim_{\beta \rightarrow \pm n\pi} \frac{ \: sin N\beta}{sin \beta}=\lim_{\beta \rightarrow \pm n\pi} \frac{\frac{d}{d\beta} (sin N\beta)}{\frac{d}{d\beta}(sin \beta)}$

$\lim_{\beta \rightarrow \pm n\pi} \frac{ \: sin N\beta}{sin \beta}=\lim_{\beta \rightarrow \pm n\pi} \frac{N cos N\beta}{cos \beta}$

$\lim_{\beta \rightarrow \pm n\pi} \frac{ \: sin N\beta}{sin \beta}=N \lim_{\beta \rightarrow \pm n\pi} \frac{ cos N\beta}{cos \beta}$

Where the value of $\lim_{\beta \rightarrow \pm n\pi} \frac{ cos N\beta}{cos \beta}=1$

$\lim_{\beta \rightarrow \pm n\pi} \frac{ \: sin N\beta}{sin \beta}=N \qquad(7)$

So the maximum intensity

$I=\frac{A^{2} \: sin^{2} \alpha}{\alpha^{2}} N^{2} \qquad(8)$

Thus, the condition for principle maxima

$sin \beta=0$

$\beta=\pm n\pi$

$(e+d)sin\theta=\pm n \lambda \qquad(9)$

For $n=0$, we get $\theta=0$ This $\theta=0$ gives the direction of zero-order principal maxima. For the value of $n=1,2,3,......$, gives the direction of first, second, third,....... order principal maxima.

Minima →

The intensity will be minimum, when $sin N\beta=0$ but $sin\beta=0$

$N\beta=\pm m\pi$

$N(e+d)sin\theta=\pm m \lambda \qquad(10)$

Where $m$ can take all integral values except $0, N,2N,3N,......$ because for these values of $m$, $sin\beta=0$ which gives the position of principal maxima.

Secondary maxima→

It is obvious from the above condition of minima, there are $(N-1)$ minima between two successive principal maxima. Hence, there are $(N-2)$ other maxima with alternative minima between two successive principal maxima. These $(N-2)$ maxima are called secondary maxima. To find the condition of secondary maxima equation $(6)$ is differentiated with respect to $\beta$ and equated to zero.

$\frac{dI}{d\beta}= \frac{A^{2}\:sin^{2}\alpha}{\alpha^{2}}2 \frac{sinN\beta}{sin\beta} \left [\frac{sin\beta . N. cosN\beta-sinN\beta . cos\beta}{sin^{2}\beta} \right ]$

$0= \frac{A^{2}\:sin^{2}\alpha}{\alpha^{2}}2 \frac{sinN\beta}{sin\beta} \left [\frac{sin\beta . N. cosN\beta-sinN\beta . cos\beta}{sin^{2}\beta} \right ]$

$N.sin\beta . cosN\beta - sinN \beta . cos\beta=0$

$\tan N\beta = N tan \beta \qquad(11)$

Now construct a right-angled triangle with the sides according to the above equation$(11)$
 Right-angled Triangle for Intensity Calculation
From the above triangle:

$sinN\beta=\frac{N tan\beta}{\sqrt{1+N^{2}tan^{2}\beta}} \qquad(12)$

Substituting the value of $sinN\beta$ from the above equation to equation (6)

$I=\frac{A^{2} \: sin^{2} \alpha}{\alpha^{2}} \frac{N^{2} tan^{2}\beta}{1+N^{2}tan^{2}\beta} \frac{ 1}{sin^{2} \beta}$

$I=\frac{A^{2} \: sin^{2} \alpha}{\alpha^{2}} \frac{N^{2}}{1+N^{2}tan^{2}\beta} \frac{ 1}{cos^{2} \beta} \qquad \left(\because tan\beta =\frac{sin\beta}{cos\beta} \right)$

$I=\frac{A^{2} \: sin^{2} \alpha}{\alpha^{2}} \frac{N^{2}}{cos^{2} \beta+N^{2}sin^{2}\beta}$

$I=\frac{A^{2} \: sin^{2} \alpha}{\alpha^{2}} \frac{N^{2}}{1- sin^{2} \beta+N^{2}sin^{2}\beta}$

$I=\frac{A^{2} \: sin^{2} \alpha}{\alpha^{2}} \frac{N^{2}}{1+(N^{2}-1) sin^{2} \beta} \qquad(12)$

Now divide the equation $(12)$ by equation $(8)$ so

$\frac{Intensity\: of\:secondary\:maxima}{Intensity\:of\:principal\:maxima}=\frac{1}{1+(N^{2}-1) sin^{2} \beta}$

It is obvious from the above equation that When $N$ increases then the intensity of secondary maxima decreases.

 Intensity distribution diagram due to a diffraction grating

### Numerical Aperture and Acceptance Angle of the Optical Fibre

Angle of Acceptance → If incident angle of light on the core for which the incident angle on the core-cladding interface equals the critical angle then incident angle of light on the core is called the "Angle of Acceptance. Transmission of light when the incident angle is equal to the acceptance angle If the incident angle is greater then the acceptance angle i.e. $\theta_{i}>\theta_{0}$ then the angle of incidence on the core-cladding interface will be less than the critical angle due to which part of incident light is transmitted into cladding as shown in the figure below Transmission of light when the incident angle is greater than the acceptance angle If the incident angle is less then the acceptance angle i.e. $\theta_{i}<\theta_{0}$ then the angle of incidence on the core-cladding interface will be greater than the critical angle for which total internal reflection takes place inside the core. As shown in the figure below Transmission of light w

### Fraunhofer diffraction due to a single slit

Let $S$ be a point monochromatic source of light of wavelength $\lambda$ placed at the focus of collimating lens $L_{1}$. The light beam is incident normally from $S$ on a narrow slit $AB$ of width $e$ and is diffracted from it. The diffracted beam is focused at the screen $XY$ by another converging lens $L_{2}$. The diffraction pattern having a central bright band followed by an alternative dark and bright band of decreasing intensity on both sides is obtained. Analytical Explanation: The light from the source $S$ is incident as a plane wavefront on the slit $AB$. According to Huygens's wave theory, every point in $AB$ sends out secondary waves in all directions. The undeviated ray from $AB$ is focused at $C$ on the screen by the lens $L_{2}$ while the rays diffracted through an angle $\theta$ are focussed at point $p$ on the screen. The rays from the ends $A$ and $B$ reach $C$ in the same phase and hence the intensity is maximum. Fraunhofer diffraction due to

### Particle in one dimensional box (Infinite Potential Well)

Let us consider a particle of mass $m$ that is confined to one-dimensional region $0 \leq x \leq L$ or the particle is restricted to move along the $x$-axis between $x=0$ and $x=L$. Let the particle can move freely in either direction, between $x=0$ and $x=L$. The endpoints of the region behave as ideally reflecting barriers so that the particle can not leave the region. A potential energy function $V(x)$ for this situation is shown in the figure below. Particle in One-Dimensional Box(Infinite Potential Well) The potential energy inside the one -dimensional box can be represented as $\begin{Bmatrix} V(x)=0 &for \: 0\leq x \leq L \\ V(x)=\infty & for \: 0> x > L \\ \end{Bmatrix}$ $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2m}{\hbar^{2}}(E-V)\psi(x)=0 \qquad(1)$ If the particle is free in a one-dimensional box, Schrodinger's wave equation can be written as: $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2mE}{\hbar^{2}}\psi(x)=0$ \$\frac{d^{2} \psi(x)}{d x