Spreading of light into the geomatric shadow of an obstacle or an aperture in the path of propagation of lgiht

If the beam of light is monochromatic, then one can see bright and dark fringes or rings or patterns, depending on the geometry of the obstacle.

Speading of light into the geomatric shadow of an obstacle or an aperture in the path of propagation of light

If the beam of light is monochromatic then one can see bright and dark fringes or rings or patterns depending on the geomatry of the obstacle.

$\lambda << W < D$

Onest of diffraction effects

if we further reduce the slit width 'W'

Path difference = $(r_2 - r_1)$

Phase difference = K $(r_2 - r_1)$

$\delta$

we treated $S_1$ and $S_2$ as a point source

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

$\Delta = a sin \theta$

Enlarged view when L is large

Due to the finite width of the slit, there is a path difference between waver emanating from any two point sources in the aperture of the slit.

The corresponding phase shift depends on $\theta$

$\Rightarrow$ The intensity of $P$ depends on $\theta$.

1. Fraunhofer Diffraction.

2. Fresnel Diffraction If the source of light and the observation screen are at large distances from the diffraction aperture, so that the wavafronts arriving at the aperture and the screen may be considered plane, then it corresponds to Fraunhofer Diffraction.

Every point $P$ on the screen corresponds to a different angle $\theta I(\theta) \equiv I(x)$.

The lens does not introduce any additional path difference (or phase difference) among the interfersing parallel set of rays.

Practical Arrangement to observe Fraunhofer Diffraction

$I(\theta) = I_o \frac {sin^2 \beta}{\beta^2}$

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

$I_0$ is the intensity for $\theta$ = 0

i.e at the point O, on th axis

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

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

$\beta = \frac {\sin\beta}{\beta}$

$I_{\theta=o} = I_0$

Single slit Diffraction: Intensity Distribution

$I_{\theta} = I_{\beta} = I_0 \frac {sin^2 \beta}{\beta^2}$

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

$\beta \rightarrow 0$

$\frac {sin \beta}{\beta} = 1$

$I (\theta = 0) = I_0$

$I_0 sin^2 \beta \times \frac {1}{\beta^2}$

$I(\theta)=I(\beta)=I \cdot \frac{\sin ^2 \beta}{\beta^2}$

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

$\Rightarrow \beta=m \pi, \quad m \neq 0$

i e for a $\sin \theta=m \lambda ; \quad m= \pm 1, \pm 2, \ldots$

Thus, the first intensity minimum will occur at $\theta_1=\sin ^{-1}\left(\frac{\lambda}{\alpha}\right)$

and at $-\theta_1=-\sin ^{-1}\left(\frac{\lambda}{a}\right)$, on either side of the central maxima, at $\theta=0$.

$\sin \theta_{\min }=\frac{\lambda}{a}$

$\lambda=500 mm =5 \times 10^{-5} cm$

a = 1 mm

$\frac{\lambda}{a}=\frac{5 \times 10^{-5}}{10^{-1}}=5 \times 10^{-4} \mathrm{rad} .$

$\sin \theta \approx \theta$

a=0-1mm

$\sin \theta=\frac{\lambda}{a}=\frac{5 \times 10^{-5}}{10^{-2}}=5 \times 10^{-3}$