$I(\theta) \equiv I(\beta)=I_0 \frac{\sin ^2 \beta}{\beta^2} ; \beta=\frac{\pi}{\lambda} a \sin \theta, I_0=I(0=0)$

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?

$\frac{d I}{d \beta}=\frac{d}{d \beta}\left[I_0 \frac{\sin ^2 \beta}{\beta^2}\right]=0$

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

Solutions:

$\beta=0$

$\beta_1=1.43 \pi$

$\beta_2=2.46 \pi$

Intensity of the first maximum,

$I_1=I_0 \frac{\sin ^2(1+43 \pi)}{(1.43 \pi)^2}=0.0496 I_0 < 5$

llly,

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

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.

$I=4 I_0 \cdot \cos ^2\left(E_2\right)$

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

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

$r_2-r_1=d \sin \theta$

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

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

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

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

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

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

$I_0 (\frac {sin^2 \beta}{\beta}). cos^2 \gamma$ 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:

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.

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 $\frac{\lambda}{a}$

a = 0.1 mm

L = 1m

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

2 l = 13 mm

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

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

2 W =$2 f \cdot \frac{\lambda}{a} \Rightarrow a=\frac{2 \times 15 \mathrm{cm} \times \lambda}{5 \mathrm{mm}}$

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

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

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

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

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

=${0.03534 \mathrm{mm}}$

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