physics-class-12-unit-10-chapter-05-optics-fringe-shift-in-the-two-hole-interferenc-equipment-l-5-9-mrqnkt3unry

### <Success style="font-size:25px"> Optics Fringe Shift In The Two Hole Interferenc Equipment L-5 </Success>
### <primary style="font-size:24px"> Fringe </primary>
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- $\Delta <<d << D$

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<footer style = "font-size: 18px"> $\rightarrow$ Optics Fringe Shift in the Two-Hole Interference Equipment $\rightarrow$ <span style = "color:blue"> Fringe </span> $\rightarrow$ Fringe $\rightarrow$ Optical Path Difference </footer>

### <Success style="font-size:25px"> Optics Fringe Shift In The Two Hole Interferenc Equipment L-5 </Success>
### <primary style="font-size:24px"> Light Waves in a Medium </primary>
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- $ \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

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<footer style = "font-size: 18px">Fringe $\rightarrow$ Optical Path Difference $\rightarrow$ <span style = "color:blue"> Light Waves in a Medium </span> $\rightarrow$ Light Waves in a Medium $\rightarrow$ A Thin Sheet in Front </footer>

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### <primary style="font-size:24px"> Light Waves in a Medium </primary>
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- 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,..

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<footer style = "font-size: 18px">Optical Path Difference $\rightarrow$ Light Waves in a Medium $\rightarrow$ <span style = "color:blue"> Light Waves in a Medium </span> $\rightarrow$ A Thin Sheet in Front $\rightarrow$ A Thin Sheet in Front </footer>

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### <primary style="font-size:24px"> A Thin Sheet in Front </primary>
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- 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)} $

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<footer style = "font-size: 18px">Light Waves in a Medium $\rightarrow$ Light Waves in a Medium $\rightarrow$ <span style = "color:blue"> A Thin Sheet in Front </span> $\rightarrow$ A Thin Sheet in Front $\rightarrow$ For the Central Fringe </footer>

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- $\delta=k_0\left[\left(r_2-r_1\right)+t(1-n)\right]= \pm N 2 \pi$  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

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<footer style = "font-size: 18px">Light Waves in a Medium $\rightarrow$ A Thin Sheet in Front $\rightarrow$ <span style = "color:blue"> A Thin Sheet in Front </span> $\rightarrow$ For the Central Fringe $\rightarrow$ Young's Interference Experiment </footer>

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### <primary style="font-size:24px"> For the Central Fringe </primary>
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- $(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.

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<footer style = "font-size: 18px">A Thin Sheet in Front $\rightarrow$ A Thin Sheet in Front $\rightarrow$ <span style = "color:blue"> For the Central Fringe </span> $\rightarrow$ Young's Interference Experiment $\rightarrow$ Double Hole vs Double Slit </footer>

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### <primary style="font-size:24px"> Young's Interference Experiment </primary>
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- 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

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<footer style = "font-size: 18px">A Thin Sheet in Front $\rightarrow$ For the Central Fringe $\rightarrow$ <span style = "color:blue"> Young's Interference Experiment </span> $\rightarrow$ Double Hole vs Double Slit $\rightarrow$ Young's Double-Slit Experiment </footer>

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### <primary style="font-size:24px"> Young's Double-Slit Experiment </primary>
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- 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

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<footer style = "font-size: 18px">Young's Interference Experiment $\rightarrow$ Double Hole vs Double Slit $\rightarrow$ <span style = "color:blue"> Young's Double-Slit Experiment </span> $\rightarrow$ Exercise-1 $\rightarrow$ Exercise-2 </footer>

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### <primary style="font-size:24px"> Exercise-1 </primary>
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- $ \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}$

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<footer style = "font-size: 18px">Double Hole vs Double Slit $\rightarrow$ Young's Double-Slit Experiment $\rightarrow$ <span style = "color:blue"> Exercise-1 </span> $\rightarrow$ Exercise-2 $\rightarrow$ Solution </footer>

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### <primary style="font-size:24px"> Exercise-2 </primary>
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- In a Young's double-slit experiment using monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where path difference is ${\lambda}$ is $K$ units. What is the intensity of light at a point where the path difference is $\lambda / 3$ ?

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<footer style = "font-size: 18px">Young's Double-Slit Experiment $\rightarrow$ Exercise-1 $\rightarrow$ <span style = "color:blue"> Exercise-2 </span> $\rightarrow$ Solution $\rightarrow$ Exarcise-3 </footer>

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### <primary style="font-size:24px"> Solution </primary>
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- $I = I_{max} cos^2 \frac{\delta}{2}$

- $ 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

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

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<footer style = "font-size: 18px">Exercise-1 $\rightarrow$ Exercise-2 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Exarcise-3 $\rightarrow$ Solution </footer>

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### <primary style="font-size:24px"> Exarcise-3 </primary>
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- In a Young's double -slit experiment, the light source used was emitting two distinct wavelengths of $440 \mathrm{nm}$ (Blue) and $660 \mathrm{nm}$ (Red). The interference fringe pattern an a screen placed at a distance $D=90 \mathrm{cm}$ showed two bright red fringes on either side of the central bright fringe. If the separation between the two slits of the double -slit aperture is $0.3 \mathrm{mm}$, what is the separation between the two bright red fringes?

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<footer style = "font-size: 18px">Exercise-2 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Exarcise-3 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

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### <primary style="font-size:24px"> Solution </primary>
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- $\rightarrow$ 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**

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<footer style = "font-size: 18px">Solution $\rightarrow$ Exarcise-3 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Exercise-4 </footer>

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### <primary style="font-size:24px"> Solution </primary>
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- Separation = $ X_1 - X_{-1}$

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

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

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

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

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<footer style = "font-size: 18px">Exarcise-3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Exercise-4 $\rightarrow$ Solution </footer>

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### <primary style="font-size:24px"> Exercise-4 </primary>
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- In a Young's double -slit arrangement with monochromatic light source, the separation between the two slits is $0.5 \mathrm{mm}$ and the screen is placed at a distance of $1 \mathrm{~m}$. When a thin transparent plastic $(n=1.5)$ sheet is placed over one of the slits, the fringe pattern is shifted to one side through a distance of $5 \mathrm{cm}$. What is the thickness of the sheet?

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Exercise-4 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>