• In Optics, it generally refers to 'Two-Beam' or 'Two-Wave' interference, which results in some perceptible and sustained 'fringe pattern:

• 'Fringe Pattern' refers to the light intensity pattern comprising of alternate bright and dark regions in the form of lines or rings.

• Formation of fringes can be explained in terms of superposition of the two waves.

• To be able to see a sustained fringe pattern,

• 1. The two waves interfering must have the same frequency (or wavelength).
• 2. There should be a constant phase difference between the two waves.

• $S, S_1$ and $S_2$ and small holes in an opaque screen

• Superposition of waves occurs at each point P

• d = difference between $S_1$ & $S_2$

• Longitudinal Section in the $x-z$ plane

• At an arbitrary point $P$,

• $\psi_1=\frac{A_1^{\prime}}{r_1} \cos \left(k r_1-\omega t\right),$

• $\psi_2=\frac{A_1^{\prime}}{r_2} \cos \left(k r_2-\omega t\right)$

• Let $A_1=\frac{A_1^{\prime}}{r_1^{\prime}}$ and $A_2=\frac{A_2^{\prime}}{r_2}$

• $\psi_{\text {resultant}}= \psi_1+\psi_2$

• = $A_1 \cos \left(k r_1-\omega t\right)+A_2 \cos \left(k r_2-\omega t\right)$

• = $A_1\left[\cos k r_1 \cos \omega t+\sin k r_1 \sin \omega t\right]$

• $+A_2\left[\cos k r_2 \cos \omega t+\sin k r_2 \sin \omega t\right]$

• (using $\cos (A-B)=\cos A \cos B+\sin A \sin B$ )

• $\psi_{\text {res. }}= \cos \omega t \left(A_1 \cos k r_1+A_2 \cos k r_2\right)+\sin \omega t\left(A_1 \sin k r_1+A_2 \sin k r_2\right)$

• Setting $\left(A_1 \cos k r_1+A_2 \cos k r_2\right)=A \cos \phi$ and $\left(A_1 \sin k r_1+A_2 \sin k r_2\right)=A \sin \phi$

• $\text { with } \tan \phi=\frac{\left(A_1 \sin k r_1+A_2 \sin k r_2\right)}{\left(A_1 \cos k r_1+A_2 \cos k r_2\right)}$

• $\psi_{\text {vas. }}=A \cos \omega t \cos \phi+A \sin \omega t \sin \phi$

• or $\psi_{\text {res. }}= A \cos {\omega t - \phi}$ at the point P

• with $A=\left(A^2 \cos ^2 \phi+A^2 \sin ^2 \phi\right)^{1 / 2}$

• Superposition of Two Waves (contd)

• Intensity at the point P $I=\left|\psi_{\text {res }}\right|^2$

  • $\quad=A^2 \cos ^2(\omega t-\phi)$

• $\left\langle\cos ^2(\omega t-\phi)\right\rangle-1 / 2$

• $\omega=2 \pi \nu \longrightarrow 10^{14}-10^{15} \mathrm{Hz} \text { for light- }$

• $\cos ^2(\omega t-\phi)$

• $<\cos ^2(\omega t-\phi)>$ time average

• = $\frac {1}{2}$

• Intensity at the point P I= $\left|\psi_{\text {res. }}\right|^2$

• $=A^2 \cos ^2(\omega t-\phi)$

• $\left\langle\cos ^2(\omega t-\phi)\right\rangle-1 / 2$

• time average

• If only the source $S_1$ war present, then intensity of the point $P$ due to $S_1$ is given by

• $I_1=\left|\psi_1\right|^2=A_1^2\left\langle\cos ^2\left(k_1 r_1-\omega t\right)\right\rangle=\underline{\frac{A_1^2}{2}}$

• parallely, Intensity at $p$ due to only $S_2$,

• $I_2=\left|\Psi_2\right|^2=\frac{A_2^2}{2} .$

• Superposition of Two Waves (contd.)

• Now,

• $A^2= \left(A_1 \cos k r_1+A_2 \cos k r_2\right)^2$

• $+\left(A_1 \sin k r_1+A_2 \sin k r_2\right)^2$

• = $A_1^2+A_2^2+2 A_1 A_2 \cos \left(k r_1-k r_2\right)$

• ie. $2 I=2 I_1+2 I_2+2 \sqrt{2 I_1} \sqrt{2 I_2} \cos \delta$

• where $\delta=k\left(r_2-r_1\right) \rightarrow$ the "phase difference" between the two waves at $P$.

• $\left(r_2-r_1\right)$ is the "path difference" between the two waves, or $I=I_1+I_2+2 \sqrt{I_1} \sqrt{I_2} \cos \delta$ "Interference Equation"

• Superposition of Two Waves (contd.)

• $\psi_{res.}= \cos \cos \left(A_1 \cos k r_1+A_2 \cos k r_2\right)$

  • $+\sin \omega t\left(A_1 \sin k r_1+A_2 \sin k r_2\right)$

• setting $\left(A_1 \cos k r_1+A_2 \cos k r_2\right)=A \cos \phi$

• $\text { and }\left(A_1 \sin k r_1+A_2 \sin k r_2\right)=A \sin \phi \text {, }$

• $\text { with } \tan \phi=\frac{\left(A_1 \sin k r_1+A_2 \sin k r_2\right)}{\left(A_1 \cos k r_1+A_2 \cos k r_2\right)}$

• $\psi_{\text {var. }}=A \cos \omega t \cos \phi+A \sin \omega t \sin \phi$

• or $\psi_{\text {res. }}=A \cos (\omega t-\phi)$ of the point $P$.

• with $A=\left(A^2 \cos ^2 \phi+A^2 \sin ^2 \phi\right)^{1 / 2}$

• $I=I_1+I_2+2 \sqrt{I_1} \sqrt{I_2} \cos \delta$

• For $\delta=0, \pm 2 \pi, \pm 4 . \pi, \ldots$.

• $I=I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2$

• For $\delta= \pm \pi, \pm 3 \pi$,

• $I=I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2$

• If $A_1=A_2, I_1=I_2=I_2$ (say),

• then

• $I_{\max }=4 I_0-\varepsilon=0, \pm 2 \pi, 3,4 \pi$

• $I_{\min }=0-\delta= \pm \pi, \pm 3 \pi, .$

& $I=2 I_0(1+\cos \delta)=1+I_0 \cos ^2\left(\frac{\delta}{2}\right)$

• Schematic of Young's

• Experimental Setup

• $S, S_1$ and $S_2$ are small holes in an opaque screen

• Superposition of waves occurs at each point P

• d = difference between $S_1$ & $S_2$

• Longitudinal Section in the $x-z$ plane

• $I(\delta)=4 I \cdot \cos ^2\left(\frac{\delta}{2}\right)$

• $\delta=k\left(r_2-r_1\right)$

• Path Difference $\left(r_2-r_1\right)$

• For a point $P$ on the $x$-axis,

• $r_2-r_1=S_2 p-S_1 p$

• $S_2 p^2=D^2+(x+d / 2)^2$

• $S_1 p^2=D^2+(x-d / 2)^2$

• $\therefore S_2 p^2-S_1 p^2=2 x d$

• $\therefore S_2 p-S_1 p=2 x d /\left(S_2 p+S_1 p\right)$

• Path difference $\left(r_2-r_1\right)=\frac{2 x d}{\left(S_1 p+S_2 p\right)}$

• In practice, $D$ : $50-100 \mathrm{~cm}, x$ : far mm d: $0.1-1 \mathrm{mm}$

• Noting that $x, d \ll D, S_1 p+S_2 p \cong 2 D$

• $\therefore$ Path Difference $r_2-r_1=\frac{x d}{D}$

• $\left(r_2-r_1\right) \propto x$, position coordinate of $P$.

• RECALL:

• $I=4 I_0 \cos ^2(\delta / 2)$

• Intensity Maxima are given by -

• $\text { Phase Difference } \delta=\frac{2vv \pi}{\lambda}\left(r_2-r_1\right)= \pm n, 2 \pi$

• with $n=0,1,2, \ldots \ldots$

• or Path Difference $\left(r_2-r_1\right)= \pm n \lambda$ '

• $n$ ' is called 'order' of the maxima.

• Intensity Minima ara given by -

• Phase Difference $\delta=\frac{2 \pi}{\lambda}\left(r_2-r_1\right)= \pm(n+1 / 2) 2 \pi$ with $n=0,1,2, \cdots$

• or Path Difference $\left(r_2-r_1\right)= \pm(n+1 / 2) x$

' + 'sign gives positions of maxima on one site of the point $\delta$=0, while. - 'sign given the other ride.

• Path Difference $\left(r_2-r_1\right)=\frac{2 x d}{\left(S_1 p+S_2 p\right)}$

• In practice, $D$ : $50-100 \mathrm{~cm}, x$ : far mm d: $0.1-1 \mathrm{mm}$

• Noting that $x, d \ll D, S_1 p+S_2 p \cong 2 D$

• $\therefore$ Path Difference $r_2-r_1=\frac{x d}{D}$

• $\left(r_2-r_1\right) \propto x$, position coordinate of $P$.

• Intensity Maxima and Minima

• RECALL:

• $I=4 I_0 \cos ^2(\delta / 2)$

• $I=4 I_0 \cos ^2(\delta / 2)$

• Intensity Maxima are given by -

• $\text { Phase Difference } \delta=\frac{1 \pi}{\lambda}\left(r_2-r_1\right)= \pm n, 2 \pi$

• with $n=0,1,2, \ldots \ldots$

• or Path Difference $\left(r_2-r_1\right)= \pm n \lambda$ '

• $n$ ' is called 'order' of the maxima.

• Intensity Minima ara given by -

• Phase Difference $\delta=\frac{2 \pi}{\lambda}\left(r_2-r_1\right)= \pm(n+1 / 2) 2 \pi$ with $n=0,1,2, \cdots$

• or Path Difference $\left(r_2-r_1\right)= \pm(n+1 / 2) x$

' + 'sign gives positions of maxima on one site of the point $\delta$=0$, while. - 'sign given the other ride.

• At the point $0,S_1 0 = S_2 0$ $\Rightarrow\left(r_2-r_1\right)=0, \rightarrow n=0$. It is a paint of maximum intensity, called "zeroth order maxima" or "Central maxima"

• $\rightarrow$ zero path difference point at nest maxima asur when it difference $\left(r_2-r_1\right)=\frac{x_1 d}{D}=\lambda ;(n-1)$

• Position of first order maximum, $x_1=\left(\frac{D}{d}\right)^\lambda$ antral, position of the order maxima is joining } $x_n=n(\frac {D}{d})^\lambda$

• Path Difference $\left(r_2-r_1\right)=\frac{2 x d}{\left(S_1 p+S_2 p\right)}$

• In practice, $D$ : $50-100 \mathrm{~cm}, x$ :o far mm d: $0.1-1 \mathrm{mm}$

• Noting that $x, d \ll D, c_1 p+S_2 p \cong 2 D$

• $\therefore$ Path Difference $r_2-r_1=\frac{x d}{D}$

• $\left(r_2-r_1\right) \propto x$, position coordinate of $P$.

• Intensity Maxima are given by -

• $\text { Phase Difference } \delta=\frac{1 \pi}{\lambda}\left(r_2-r_1\right)= \pm n, 2 \pi$

• with $n=0,1,2, \ldots \ldots$

• or Path Difference $\left(r_2-r_1\right)= \pm n \lambda$ '

• $n$ ' is called 'order' of the maxima.

• Intensity Minima ara given by -

• Phase Difference $\delta=\frac{2 \pi}{\lambda}\left(r_2-r_1\right)= \pm(n+1 / 2) 2 \pi$ with $n=0,1,2, \cdots$

• or Path Difference $\left(r_2-r_1\right)= \pm(n+1 / 2) x$

' + 'sign gives positions of maxima on one site of the paint $\delta$=0$, while. - 'sign given the other ride.

• Separation between adjacent maxima $=x_{n+1}-x_n$ or

• $\beta=(n+1) \frac{D}{d} \lambda-n \frac{D}{d} \lambda=\left(\frac{D}{d}\right)^\lambda$

• Note that $\beta \propto D$ and $\propto \frac{1}{d}$ at a given wavelength.

• Typical: $D=100 \mathrm{~cm}, d=0.3 \mathrm{mm}, \lambda=600 \mathrm{nm}$

• $ \beta=\frac{100 \times 10^{-2} \times 600 \times 10^{-9}}{0.3 \times 10^{-8}} \mathrm{m}=2 \mathrm{mm}

• $

• Similarly, positions of minima are given by -

• $r_2-r_1=\frac{x_m d}{D}= \pm(m+1 / 2) \lambda, \quad m=0,1,2 ;$

• Thus, $\quad x_m=(m+1 / 2)\left(\frac{D}{d}\right)^\lambda$

• $m=0$ gives position of first minima, $m=1$ gives position of second minima

• Path Difference at a point $Q$ in the $x-y$ plane

• Coordinates of positions Q,$S_1 and S_2$

• Q : (x, y, 0)

• $S_1 : (\frac{d}{2}, 0, -D)$

• $S_2 : (\frac {-d}{2}, 0, -D)$

• Path Difference $\left(r_2-r_1\right)=S_2 Q-s, Q$.

• Path Difference at the point $Q=S_2 Q - S_1 Q,-\Delta$, say

• ${[(x+d / 2)^2+y^2+D^2]^{1 / 2}}$ -${(x-d / 2)^2+y^2+D^2]^{1 / 2}-\Delta}$

• or $(x+d / 2)^2+y^2+D^2= [{\Delta+ [(x-d / 2)^2+y^2+D^2]^{1 / 2}}]^2$

• on simplification, we get

• $\left(d^2-\Delta^2\right) x^2-\Delta^2 y^2=\Delta^2\left[D^2+\frac{1}{2}\left(A^2-\Delta^2\right)\right]$

• for a fixed value of $\Delta$, the above equation is of the form

• $\quad \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

• Equation of a Hyperbola.

• where $a$ & $b$ are constants.

• For different values of $\Delta$, we get different Hyperbola.

• ie. the locus of all points with constant path differing are hyperbola.

• d $\sim 0.1 - 1 mm \sim mn$

• $\Delta \rightarrow \lambda , 2\lambda, 3\lambda,$ for $\lambda$

• 600 nm $\rightarrow 0.6 \mu m \sim \mu m$

• d >> $\Delta$

• Path Difference at the point $Q=S_2 Q - S_1 Q,-\Delta$, say

• ${[(x+d / 2)^2+y^2+D^2]^{1 / 2}}$

-${(x-d / 2)^2+y^2+D^2]^{1 / 2}-\Delta}$

• or $(x+d / 2)^2+y^2+D^2= [{\Delta+ [(x-d / 2)^2+y^2+D^2]^{1 / 2}}]^2$

• on simplification, we get

• $\left(d^2-\Delta^2\right) x^2-\Delta^2 y^2=\Delta^2\left[D^2+\frac{1}{2}\left(A^2-\Delta^2\right)\right]$

• for a fixed value of $\Delta$, the above equation is of the form

• $\quad \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

• Equation of a Hyperbola.

• where $a$ & $b$ are constants.

• For different values of $\Delta$, we get different Hyperbola.

• ie. the locus of all points with constant path differing are hyperbola.

• Path Difference at the point $Q=S_2 Q - S_1 Q,-\Delta$, say

• ${[(x+d / 2)^2+y^2+D^2]^{1 / 2}}$

-${(x-d / 2)^2+y^2+D^2]^{1 / 2}-\Delta}$

• or $(x+d / 2)^2+y^2+D^2= [{\Delta+ [(x-d / 2)^2+y^2+D^2]^{1 / 2}}]^2$

• on simplification, we get

• $\left(d^2-\Delta^2\right) x^2-\Delta^2 y^2=\Delta^2\left[D^2+\frac{1}{2}\left(A^2-\Delta^2\right)\right]$

• for a fixed value of $\Delta$, the above equation is of the form

• $\quad \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

• Equation of a Hyperbola.

• where $a$ & $b$ are constants.

• For different values of $\Delta$, we get different Hyperbola.

• ie. the locus of all points with constant path differing are hyperbola.

• If $\Delta=n \lambda$, then the hyperbola would comprise of all points with intensity maxima, and if $\Delta=\left(n+ \frac {1}{2}\right) \lambda$, then the hyperbola would comprise of all points with intensity minima

• $\Rightarrow$ We will see alternate bright and dark hyperbola on the screen.

• These are called "interference fringes", and the pattern on the screen is called "fringe pattern"

• If the locus of all points with a constant path difference happens to he circles, we would see alternate bright and dark rings" or "circular fringes" on a screen.

• Path Difference $(r_2 - r_2)$ = $\frac {2xd}{(S_1 P + S_2 P)}$

• In practice D : 50-100 cm, x = few mm

• d : 0.1 - 1 mm

• Noting that x, d << D , $S_1 P + S_2 P$ = 2D

• Path difference $r_2 - r_1 = \frac {xd}{D}$

• $(r_2 - r_1) \propto x$ position coordinates of P.

• Interference Fringes in Young's Experiment

• $\left(d^2-\Delta^2\right) x^2-\Delta^2 y^2=\Delta^2\left[D^2+\frac{1}{4}\left(d^2-\Delta^2\right)\right]$

• $\quad x={\frac{\Delta}{\left(d^2-\Delta^2\right)^{1 / 2}}\left[D^2+y^2+\frac{1}{4}\left(d^2-\Delta^2\right)\right]^{1 / 2}}$

• In practice, $y \ll D$ ( $y-$ few $\mathrm{mm}$ to $\mathrm{cm}, D \sim 100 \mathrm{~cm})$

• $\therefore y^2$ can be neglected in comparison to $D^2$.

• or x=$\frac{\Delta}{\left(d^2-\Delta\right)^{1 / 2}}\left[D^2+\frac{1}{4}\left(d^2-\Delta^3\right)\right]^{1 / 2}$

• $\therefore e . x=$ constant, for each (fixed) value of $\Delta$

• $\Rightarrow$ locus of constant path difference are straight lines parallel to the $y$-axis ($\perp$ to $x$-air)

• $\Rightarrow$ straight line interference fringes II

• The separation between adjacent bright (or dark) fringes is called "fringe width".

• Recall: $\beta=\left(\frac{D}{d}\right)^\lambda$ - separation between maxima on the x-axis

• $x ={\frac{\Delta}{\left(d^2-\Delta^2\right)^{1 / 2}}}\left[D^2+\frac{1}{4}\left(d^2-\Delta^2\right)\right]^{1 / 2}$

• $\approx \frac{\Delta D}{d} \text {, as } \Delta \ll d \ll D$

• For $\Delta=0, h^2=0$ - constant

• $\rightarrow y$-axis is the locus of all bright points

• ie. the "central fringe".

• $\Delta-\lambda, \quad x_1=\frac{\lambda D}{d}$, for the first order bright fringe

• $\Delta=2 \lambda, x_2= \frac {2 \lambda D} {d}, 2^{nd}$ bright fringe,

• Fringe width $\beta$ = $x_2 - x_1$ = $\frac {\lambda D}{d}$