physics-class-12-unit-10-chapter-07-diffraction-patterns-due-to-a-single-slit-anda-circular-l-7-9-vcwflttbah0

### <Success style="font-size:25px"> Diffraction Patterns Due To A Single Slit Anda Circular L-7 </Success>
### <primary style="font-size:24px"> Single-Slit Diffraction </primary>
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- $I(\theta) \equiv I(\beta)=I_0 \frac{\sin ^2 \beta}{\beta^2} ; \beta=\frac{\pi}{\lambda} a \sin \theta, I_0=I(0=0) $

-  Intensity Minima : $\beta=m \pi, m= \pm 1,=2, \cdots $
 
- $\rightarrow \sin \theta_{\min }=m \frac{\lambda}{a}, m= \pm 1, \pm 2,..... $

- $ \frac{\lambda}{a} \ll 1 \rightarrow \sin \theta \approx \theta $

-  or  $\theta_{\min } = m \frac{\lambda}{a}, m= \pm 1,\pm2 .... $
 
-  What about the Intensity Maxima?

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<footer style = "font-size: 18px"> $\rightarrow$ Diffraction Pattern due to a Single Slit and a Circular $\rightarrow$ <span style = "color:blue"> Single-Slit Diffraction </span> $\rightarrow$ Intensity Maxima $\rightarrow$ Intensity Maxima </footer>

### <Success style="font-size:25px"> Diffraction Patterns Due To A Single Slit Anda Circular L-7 </Success>
### <primary style="font-size:24px"> Intensity Maxima </primary>
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- $\frac{d I}{d \beta}=\frac{d}{d \beta}\left[I_0 \frac{\sin ^2 \beta}{\beta^2}\right]=0$

- ie. $\frac{1}{\beta^2}\left[2 \sin \beta \cos \beta-\frac{2}{\gamma^2} \sin ^2 \beta\right]=0$ gives $\tan \beta=\beta$

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<footer style = "font-size: 18px">Diffraction Pattern due to a Single Slit and a Circular $\rightarrow$ Single-Slit Diffraction $\rightarrow$ <span style = "color:blue"> Intensity Maxima </span> $\rightarrow$ Intensity Maxima $\rightarrow$ Single Slit Diffraction Experiment </footer>

### <Success style="font-size:25px"> Diffraction Patterns Due To A Single Slit Anda Circular L-7 </Success>
### <primary style="font-size:24px"> Intensity Maxima </primary>
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- **Solutions:**

- $ \beta=0 $

- $ \beta_1=1.43 \pi $

- $ \beta_2=2.46 \pi$

- Intensity of the first maximum,

- $I_1=I_0 \frac{\sin ^2(1+43 \pi)}{(1.43 \pi)^2}=0.0496 I_0 < 5% of I_0!$
- llly,

- $I_2=I_0 \frac{\sin ^2(2.46 \pi)}{(2.46 \pi)^2}=0.0168 I_0$

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<footer style = "font-size: 18px">Single-Slit Diffraction $\rightarrow$ Intensity Maxima $\rightarrow$ <span style = "color:blue"> Intensity Maxima </span> $\rightarrow$ Single Slit Diffraction Experiment $\rightarrow$ Double Slit Experiment </footer>

### <Success style="font-size:25px"> Diffraction Patterns Due To A Single Slit Anda Circular L-7 </Success>
### <primary style="font-size:24px"> Single Slit Diffraction Experiment </primary>
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- Diffraction pattern on the screen

- $\frac{2 l}{L}=\frac{\lambda}{a}-\left(-\frac{\lambda}{a}\right)=\frac{2 \lambda}{a} \Rightarrow \lambda=l^.\frac{a}{L}$

- Measure the separation ' $2 l$ ' (using a graph paper for the screen), ' $a$ '- using a travelling microscope, and' $L$ ' by using a scale to determine the wavelength of the laser. - A standard experiment for undergraduates.

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<footer style = "font-size: 18px">Intensity Maxima $\rightarrow$ Intensity Maxima $\rightarrow$ <span style = "color:blue"> Single Slit Diffraction Experiment </span> $\rightarrow$ Double Slit Experiment $\rightarrow$ Double Slit Experiment </footer>

### <Success style="font-size:25px"> Diffraction Patterns Due To A Single Slit Anda Circular L-7 </Success>
### <primary style="font-size:24px"> Double Slit Experiment </primary>
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- $ I=4 I_0 \cdot \cos ^2\left(E_2\right) $

- $ \delta=k\left(r_2-r_1\right) $

- $ k=\frac{2 \pi}{\lambda} $

- $ r_2-r_1=d \sin \theta $

- $ \delta=\frac{2 \pi}{\lambda} \cdot d \sin \theta $

- For large D

- Taking into account the finite size of the slits $S_1$ and $S_2$, the intensity distribution on the screen is given by

- $ I(\theta)=I_0 \frac{\sin ^2 \beta}{\beta^2}. \cos ^2 \gamma $

- $ \beta=\frac{\pi}{\lambda} a \sin \theta $

- $ \gamma=\frac{\pi}{\lambda} d \sin \theta$

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<footer style = "font-size: 18px">Intensity Maxima $\rightarrow$ Single Slit Diffraction Experiment $\rightarrow$ <span style = "color:blue"> Double Slit Experiment </span> $\rightarrow$ Double Slit Experiment $\rightarrow$ Double slit Diffraction Pattern </footer>

### <Success style="font-size:25px"> Diffraction Patterns Due To A Single Slit Anda Circular L-7 </Success>
### <primary style="font-size:24px"> Double Slit Experiment </primary>
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- I = $I_o \frac {Sin^2 \beta }{\beta^2} . cos^2 \gamma$

- $\gamma = \frac{\pi}{\lambda} . sin \theta$

- $\beta = \frac{\pi}{\lambda} a. sin \theta$

- a<<d

- $d \sim 1 mm$

- 0.1 \- 0.2 mm

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<footer style = "font-size: 18px">Single Slit Diffraction Experiment $\rightarrow$ Double Slit Experiment $\rightarrow$ <span style = "color:blue"> Double Slit Experiment </span> $\rightarrow$ Double slit Diffraction Pattern $\rightarrow$ Double slit Diffraction Pattern </footer>

### <Success style="font-size:25px"> Diffraction Patterns Due To A Single Slit Anda Circular L-7 </Success>
### <primary style="font-size:24px"> Double slit Diffraction Pattern </primary>
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- $I_0 (\frac {sin^2 \beta}{\beta}). cos^2 \gamma $ for, d = 4a

- If $d=4.5 a$, the $5^{th}$ minima of $\cos ^2 \gamma($ ie. $\sin \theta=4.5 \lambda/ d)$ would coincide with the first diffraction minima at $\lambda / a$, and there would be no" missing secondary maxima:

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<footer style = "font-size: 18px">Double Slit Experiment $\rightarrow$ Double Slit Experiment $\rightarrow$ <span style = "color:blue"> Double slit Diffraction Pattern </span> $\rightarrow$ Double slit Diffraction Pattern $\rightarrow$ Double Slit Diffraction Pattern </footer>

### <Success style="font-size:25px"> Diffraction Patterns Due To A Single Slit Anda Circular L-7 </Success>
### <primary style="font-size:24px"> Double slit Diffraction Pattern </primary>
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- If a << d than there will be large number of secondary maxima between $sin \theta \approx \theta = \pm \lambda / a$

- Fringes of nearly equal intensity.

- The double-slit interference fringes.

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<footer style = "font-size: 18px">Double Slit Experiment $\rightarrow$ Double slit Diffraction Pattern $\rightarrow$ <span style = "color:blue"> Double slit Diffraction Pattern </span> $\rightarrow$ Double Slit Diffraction Pattern $\rightarrow$ Linear Width of the Central Fringe </footer>

### <Success style="font-size:25px"> Diffraction Patterns Due To A Single Slit Anda Circular L-7 </Success>
### <primary style="font-size:24px"> Double Slit Diffraction Pattern </primary>
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<footer style = "font-size: 18px">Double slit Diffraction Pattern $\rightarrow$ Double slit Diffraction Pattern $\rightarrow$ <span style = "color:blue"> Double Slit Diffraction Pattern </span> $\rightarrow$ Linear Width of the Central Fringe $\rightarrow$ Example-1 </footer>

### <Success style="font-size:25px"> Diffraction Patterns Due To A Single Slit Anda Circular L-7 </Success>
### <primary style="font-size:24px"> Linear Width of the Central Fringe </primary>
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- Linear width of the central fringe on the focal plane of the lens

- If the point P on the screen corresponds to the first intensity minima, then $sin\theta = \frac{\lambda }{a}$

- since $ \frac {\lambda }{a}$ << 1, $sin\theta \approx tan\theta \approx \theta$,

- $\frac {W}{F}= \frac{\lambda}{a}$

- The linear width of the central fringes on the screen

- ie. 2W = 2F $\frac{\lambda}{a}$

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<footer style = "font-size: 18px">Double slit Diffraction Pattern $\rightarrow$ Double Slit Diffraction Pattern $\rightarrow$ <span style = "color:blue"> Linear Width of the Central Fringe </span> $\rightarrow$ Example-1 $\rightarrow$ Solution-1 </footer>

### <Success style="font-size:25px"> Diffraction Patterns Due To A Single Slit Anda Circular L-7 </Success>
### <primary style="font-size:24px"> Example-1 </primary>
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- In a single-slit diffraction experiment, a parallel beam of laser light is incident normally on a long narrow slit of width $0.1 \mathrm{mm}$. The diffraction pattern is observed on a screen placed at a distance of $1 \mathrm{m}$, on the other side of the slit. If the separation between the first intensity minima on either side of the central maximum is $13 \mathrm{mm}$ (as observed on the screen), determine the wavelength of the laser.

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<footer style = "font-size: 18px">Double Slit Diffraction Pattern $\rightarrow$ Linear Width of the Central Fringe $\rightarrow$ <span style = "color:blue"> Example-1 </span> $\rightarrow$ Solution-1 $\rightarrow$ Example-2 </footer>

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### <primary style="font-size:24px"> Solution-1 </primary>
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- a = 0.1 mm

- L = 1m

- $\theta = \frac{\lambda}{a}$

- 2 l = 13 mm

- $tan \theta = \frac{l}{L} = \frac{(13/2) \ times 10{-3} m}{1m} = \frac {\lambda}{0.1 \times 10{-3}}$

- $\lambda = \frac {0.1 \times 10{-3} \ 6.5 \times 10{-3}}{1m}= 0.65 \times 10{-6}m$

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<footer style = "font-size: 18px">Linear Width of the Central Fringe $\rightarrow$ Example-1 $\rightarrow$ <span style = "color:blue"> Solution-1 </span> $\rightarrow$ Example-2 $\rightarrow$ Solution-2 </footer>

### <Success style="font-size:25px"> Diffraction Patterns Due To A Single Slit Anda Circular L-7 </Success>
### <primary style="font-size:24px"> Example-2 </primary>
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- In a singla-slit Fraunhofer diffraction experiment using Sodium lamp $(\lambda=589 \mathrm{nm})$, the separation between the two first minima $C$ on either side of the central maximum) is found to be $5 \mathrm{mm}$. If the observation screen was placed on the focal plane of a convex lens of focal length $15 \mathrm{cm}$, determine the slit width.

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<footer style = "font-size: 18px">Example-1 $\rightarrow$ Solution-1 $\rightarrow$ <span style = "color:blue"> Example-2 </span> $\rightarrow$ Solution-2 $\rightarrow$ Diffraction by a Circular Aperture </footer>

### <Success style="font-size:25px"> Diffraction Patterns Due To A Single Slit Anda Circular L-7 </Success>
### <primary style="font-size:24px"> Solution-2 </primary>
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- 2 W =$2 f \cdot \frac{\lambda}{a} \Rightarrow a=\frac{2 \times 15 \mathrm{cm} \times \lambda}{5 \mathrm{mm}}$

- $5 \mathrm{mm} \quad 15 \mathrm{cm}$.

- $ a=\frac{2 \times 315 \times 10^{-2} \times 589 \times 10^{-9}}{5 \times 10^{-3}}$

- $\frac{6 \times 589 \times 10^{-8}}{3534 \times 10^{-8} \mathrm{m}}$

- ${35.34 \times 10^{-6} \mathrm{m}}$,

- =${35.34 \mu \mathrm{m}}$

- =${0.03534 \mathrm{mm}} $

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<footer style = "font-size: 18px">Solution-1 $\rightarrow$ Example-2 $\rightarrow$ <span style = "color:blue"> Solution-2 </span> $\rightarrow$ Diffraction by a Circular Aperture $\rightarrow$ Airy Pattern </footer>

### <Success style="font-size:25px"> Diffraction Patterns Due To A Single Slit Anda Circular L-7 </Success>
### <primary style="font-size:24px"> Airy Pattern </primary>
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- The intersity distribution due to fraunhofer diffraction by a circular aperture is given by

- I = $I_0 [\frac {2 I_1 (v)}{v}]^2$,

- v= $\frac{\pi}{\lambda}.2a .sin \theta$

- $I_1$ (v) is Bessel function of the first order.

- 86% of the energy is contained within the Airy disk

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<footer style = "font-size: 18px">Solution-2 $\rightarrow$ Diffraction by a Circular Aperture $\rightarrow$ <span style = "color:blue"> Airy Pattern </span> $\rightarrow$ Thank You $\rightarrow$  </footer>