Interference occurs when two coherent waves of same frequency but different phase superpose. \begin{align} &E_1=E_{01}\sin(kx-\omega t)\\ &E_2=E_{02}\sin(kx-\omega t+\delta)\nonumber\\ &E=E_1+E_2=E_0\sin(kx-\omega t+\epsilon) \nonumber \\ & E_0=\sqrt{{E_{01}}^2 + {E_{02}}^2 + 2 E_{01} E_{02}\cos\delta} \nonumber\\ &\tan\epsilon=\frac{E_{02\sin\delta}}{E_{01}+E_{02}\cos\delta}\nonumber \end{align}
The path difference between interferening waves in a Young's double slit experiment is given by \begin{align} \Delta x=\frac{dy}{D} \end{align}
The phase difference is given by \begin{align} \delta=\frac{2\pi}{\lambda}\Delta x \end{align}
The conditions for constructive and destructive interference are: for integer $n$, \begin{align} \delta=\left\{\begin{array}{ll} 2n\pi, & \text{constructive};\\ (2n+1)\pi, & \text{destructive}, \end{array}\right.\nonumber \end{align} \begin{align} \Delta x=\left\{\begin{array}{ll} n\lambda, & \text{constructive}; \\ \left(n+\tfrac{1}{2}\right)\lambda, & \text{destructive} \end{array}\right.\nonumber \end{align}
The inetensity is given by \begin{align} I&=I_1+I_2+2\sqrt{I_1I_2}\cos\delta,\nonumber\\ I_\text{max}&=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2,\\ I_\text{min}&=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2 \end{align} If $I_1=I_2=I_0$ then \begin{align} I&=4I_0\cos^2\tfrac{\delta}{2},\\ I_\text{max}&=4I_0, \\ I_\text{min}&=0 \end{align}
The fringe width is given by \begin{align} w=\frac{\lambda D}{d} \end{align}
The ptical path is given by \begin{align} \Delta x^\prime=\mu\Delta x \end{align}
Problem (IIT JEE 2012): Young's double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe width recorded are $\beta_G,$ $\beta_R$, and $\beta_B,$ respectively. Then,
Solution: The fringe width $\beta$ is related to the wavelength $\lambda$ by, $\beta={\lambda D}/{d}$. In the visible spectrum $\lambda_B<\lambda_G <\lambda_R$ (VIBGYOR), and hence $\beta_R>\beta_G>\beta_B$.
Problem (IIT JEE 2000): In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other, then in the interference pattern,
Solution: Let the intensity of light passing through the two slits be $I_1$ and $I_2$. The intensity at a point having a phase difference $\delta$ is given by, \begin{align} I=I_1+I_2+2\sqrt{I_1I_2}\cos\delta. \end{align} The above equation gives the intensity at maxima ($\delta=0$) and minima ($\delta=\pi$) as, \begin{align} &I_\text{max}=I_1+I_2+2\sqrt{2I_1I_2},\\ &I_\text{min}=I_1+I_2-2\sqrt{I_1I_2}. \end{align} When the slits are of same width, $I_1=I_0$, $I_2=I_0$, $I_\text{max}=4I_0$ and $I_\text{min}=0$. When the width of one of the slit is doubled, \begin{align} I_1&=I_0,\\ I_2&=2I_0,\\ I_\text{max}&=(3+2\sqrt{2})I_0=5.83I_0\\ I_\text{min}&=(3-2\sqrt{2})I_0=0.17I_0. \end{align} The readers are encouraged to show that both $I_\text{max}$ and $I_\text{min}$ increase if $I_2=xI_0$ for some $x>1$.