Interference Of Waves Of Intensity $I_{1}$ and $I_{2}$ :

Resultant intensity,

$I=I_1+I_2+2\sqrt{I_1I_2}\hspace{2mm} \cos(\Delta\phi)$

where, $\Delta \phi=$ phase difference.

For Constructive Interference :

$I_{\max }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}$

For Destructive interference :

$ I_{\min }=\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}$

If sources are incoherent : $I=I_1+I_2$, at each point.

YDSE:

Path difference, $\Delta p=S_{2} P-S_{1} P=d \sin \theta$

$\begin{array}{ll}\text { if } & d<D D \quad=\frac{d y}{D} \\ \text { if } & y<D\end{array}$

for maxima, $\Delta p=n \lambda \quad \Rightarrow \quad y=n \beta \quad n = 0, \pm 1 , \pm 2, \ldots$

for minima

$ \begin{aligned} & \Delta p=\quad \Delta p= \begin{cases}(2 n-1) \frac{\lambda}{2} & n=1,2,3 \ldots \\ (2 n+1) \frac{\lambda}{2} & n=-1,-2,-3 \ldots \end{cases} \\ \\ & \Rightarrow \quad y= \begin{cases}(2 n-1) \frac{\beta}{2} & n=1,2,3 \ldots \\ (2 n+1) \frac{\beta}{2} & n=-1,-2,-3 \ldots \end{cases} \end{aligned} $

where, fringe width $\beta=\frac{\lambda D}{d}$

Here, $\lambda=$ wavelength in medium.

Highest order maxima :

$ \mathrm{n}_{\max }=\left[\frac{\mathrm{d}}{\lambda}\right]$

Total number of maxima $=2 \mathrm{n}_{\max }+1$

Highest order minima :

$\mathrm{n}_{\max }=\left[\frac{\mathrm{d}}{\lambda}+\frac{1}{2}\right]$

Total number of minima $=2 \mathrm{n}_{\max }$.

Intensity on Screen :

$ I=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}} \cos (\Delta \phi)$

where, $\Delta \phi=\frac{2 \pi}{\lambda} \Delta p$

If $I_1=I_2,\quad{I= I_1} \cos^2(\frac{\Delta\phi}{2})$

YDSE with two wavelengths $\lambda_{1} $ and $ \lambda_{2}$:

The nearest point to central maxima where the bright fringes coincide:

$y=n_{1} \beta_{1}=n_{2} \beta_{2}$

The nearest point to central maxima where the two dark fringes coincide, $y=\left(n_ {1}-\frac{1}{2}\right) \beta_{1}= \left(n_ {2}-\frac{1}{2} \right) \beta_{2}$

Optical Path Difference

$\Delta \mathrm{p}_{\mathrm{opt}}=\mu \Delta \mathrm{p} $

$\Delta \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{p}=\frac{2 \pi}{\lambda_{\text {vacuum }}} \Delta \mathrm{p}_{\text {opt. }} . $

$\Delta=(\mu-1) \mathrm{t} . \frac{\mathrm{D}}{\mathrm{d}}=(\mu-1) \mathrm{t} \frac{\mathrm{B}}{\lambda} .$

YDSE With Oblique Incidence

In YDSE, ray is incident on the slit at an inclination of $\theta_{0}$ to the axis of symmetry of the experimental set-up

We obtain central maxima at a point where, $\Delta p=0$.

$\text { or } \quad \theta_{2}=\theta_{0} \text {. }$

This corresponds to the point $\mathrm{O}^{\prime}$ in the diagram.

Hence we have path difference.

$ \Delta p = \begin{cases} d(\sin \theta_0 + \sin \theta) & \text{for points above } O \\ d(\sin \theta_0 - \sin \theta) & \text{for points between } O \text{ and } O’ \\ d(\sin \theta - \sin \theta_0) & \text{for points below } O' \end{cases} $

Thin-Film Interference

For interference in reflected light $ 2 \mu \mathrm{d}$

$= \begin{cases}n \lambda & \text { for destructive interference } \\ \left(n+\frac{1}{2}\right) \lambda & \text { for constructive interference }\end{cases}$

For interference in transmitted light $\quad 2 \mu \mathrm{d}$

$= \begin{cases}n \lambda & \text { for constructive interference } \\ \left(n+\frac{1}{2}\right) \lambda & \text { for destructive interference }\end{cases}$

Polarisation:

$\mu=\tan \theta$

Where $\theta$ is brewster’s angle

$\theta \rho+\theta_{r}=90^{\circ}$ (reflected and refracted rays are mutually perpendicular.)

Law of Malus

$I = I_0 \cos^2(\theta) = KA^2\cos^2(\theta)$

Optical Activity

$[\alpha]_{t}^{\lambda}{ }^{\circ} \mathrm{C}=\frac{\theta}{\mathrm{L} \times \mathrm{C}}$

$\theta=$ rotation in length $L$ at concentration $C$.

Diffraction

• $\quad a \sin \theta=(2 m+1) / 2$ for maxima. where $m=1,2,3 \ldots \ldots$

• $\quad \sin \theta=\frac{m \lambda}{a}, m= \pm 1, \pm 2, \pm 3 \ldots \ldots \ldots$. for minima.

• $\quad$ Linear width of central maxima $=\frac{2 \mathrm{~d} \lambda}{\mathrm{a}}$

• $\quad$ Angular width of central maxima $=\frac{2 \lambda}{a}$

• $\quad I=I_{0}\left[\frac{\sin \beta / 2}{\beta / 2}\right]^{2}$ where $\beta=\frac{\pi a \sin \theta}{\lambda}$

Resolving power:

$\mathrm{R}=\frac{\lambda}{\lambda_{2}-\lambda_{1}}=\frac{\lambda}{\Delta \lambda}$

where, $\lambda=\frac{\lambda_{1}+\lambda_{2}}{2}, \quad \Delta \lambda=\lambda_{2}-\lambda_{1}$

Davisson–Germer experiment:

The Davisson–Germer experiment was a 1923-27 experiment by Clinton Davisson and Lester Germer at Western Electric (later Bell Labs), in which electrons, scattered by the surface of a crystal of nickel metal, displayed a diffraction pattern.