$S_1$ and $S_2$ are in phase

$(r_2 - r_1) = n\lambda$

Optical path difference

$\psi=\frac{A}{r} \cos (k r-\omega t)$ - spherical Wave

$\psi=A \cos (k x-\omega t)$ - Plane Wave proprgating in $x$-direction

$k=\frac{2 \pi}{\lambda}, \lambda$ - wavelength in the medium

In a medium of refractive index $n$,

$\lambda=\frac{\lambda_0}{n} \rightarrow k=k_0 n, \lambda_0$- free space Wavelength

$\lambda_{\text {air }}=\frac{\lambda_0}{n_{\text {air }}}=\frac{\lambda_0}{1.000 \mathrm{~s}} \approx \lambda_0$

$\therefore$ We take

$k_{\text {air }}=k_0=\frac{2 \pi}{\lambda_0}$

$k_{\text {medium }}=k_0 n$

Medium referes to glass, water. mica, plastic,..

A thin sheet in front of $S_1$

Thickness of sheet - ' $t$ ' R. tractive index - ' $n$ '

Phase difference at point $P$ is -

$\delta=k_0\left[r_2-\left(r_1-t\right)\right]-k_0 n t$

$=k_0\left(r_2-r_1\right)+{k_0 t(1-n)}$

$\delta=k_0\left[\left(r_2-r_1\right)+t(1-n)\right]= \pm N 2 \pi$ for Bright Fringe N = 0,1,2,3..............

$\Rightarrow\left(r_2-r_1\right)+t(1-n)= \pm N \lambda_0$

$\Rightarrow(r_2-r_1)$ = t(n-1) for the central bright fringe

$(r_2-r_1) = t(n-1)$

i.e. $\frac{x}{D} d = t(n-1) \Rightarrow x=\frac{D}{d} t(n-1)$

$x \neq 0$, represents the "Fringe Shift".

Example: $t=10 \mu \mathrm{m}, n=1.5, d=1 \mathrm{mm}, D=1 \mathrm{m}$

Fringe Shift =$\frac{1 \mathrm{m} \times 10 \times 10^{-6} \mathrm{m} \times 0.5}{10^{-3} \mathrm{m}} = 5 \times 10^{-3} \mathrm{m}=5 \mathrm{mm}$

Note- The shift in the central fringe is independant of the wavelength of light.

Two Important Applications of Young's Interference Experiment

1. To determine the wavelength $(\lambda)$ of a monochromatic source by measuring the fringe width: $\lambda=\frac{\beta d}{D}$

2. To determine the thickness $(t)$ of a thin transparent sheet by measuring the fringe shift, $\Delta x$

t = $\frac {\Delta x}{n-1}. \frac{d}{D}$

n = Refractive index of sheet

In a young's double-slit experiment the slits are separated by 0.28mm, and the screen is placed 1.4m away. Distance between the central bright fridge and the fourth bright fringe is 1.2cm. Determine the wavelength of light need in the experiment.

Young's Double slit experiment

$\lambda=? \quad 4 \beta=1.2 \mathrm{cm} \quad \beta=0.3 \mathrm{cm} .$

$\lambda = \frac{\beta d}{D} = \frac{0.3 \times 0.028} {140cm} = \frac {280 \times 10^{-4} \times 0.3}{140}$

$0.6 \times 10^{-4}$ cm

$\lambda=600 \mathrm{mm}$

$I = I_{max} cos^2 \frac{\delta}{2}$

$I = I_{max} \frac{1}{4} = \frac {K}{4}$

$\delta = \frac {2\pi}{\lambda}{\lambda}{3}$

$\frac {\delta}{2} = \frac {\pi}{3} = 60^0$ ,{$I_{max} = K$}

Path differ = $\frac {\lambda}{3}$

I = $4 I_0 cos^2 \frac{\delta}{2}$

$\delta = k_0 x$ path diff

$\delta = \frac {2\pi}{\lambda} \times \lambda = 2\pi$

$\rightarrow$ At $x=0$, both colours will satisfy the condition for maxima, as path difference no $\Rightarrow$ bright fringe (red + blue)

$\rightarrow$ We will see a bright red fringe at any position $x$ if -

(i) the red colour satisfies the condition for maxima, viz path difference at x $=n \lambda_{red}$

AND (ii) the blue colour satisfies the condition for minima, viz. path difference at x $= (m+\frac{1}{2}) \lambda_{blue}$

NOTE -

$\lambda_{red}=(1+1 / 2) \lambda_{blue}$

In this exercise

Separation = $X_1 - X_{-1}$

= 2 $\frac {D}{d} \lambda_{red}$

= $2 \times \frac {900}{0.3} \times 660 nm$ = 3.96 mm

$x_1 = \frac{D}{d} \lambda_{red}$

$x_{-1} = -\frac{D}{d} \lambda_{red}$

Recall t(n-1) = $\Delta x \frac{d}{D}$

Additional optical path difference due to the sheet = fringe shift $\times \frac {d}{D}$

$r_2 - r_1 = x \frac {d}{D}$

t (n-1), (+ $\Delta x {d}{D})$

t = $\frac{\Delta x}{n-1}. \frac {d}{D}$

d = 0.5 mm, D = 1000mm

n = 1.5 , $\Delta x = 50 mm$

t = $\frac {0.5 mm}{(1.5 -1)} \times \frac {50mm}{1000mm}$ mm

= $50 \times 10{-3}$mm = 50 $\mu m$