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.