Diffraction

DIFFRACTION

1) Diffraction is the bending or spreading of waves that encounter an object ( a barrier or an opening) in their path.

2) In Fresnel class of diffraction, the source and/or screen are at a finite distance from the aperture.

3) In Fraunhoffer class of diffraction, the source and screen are at infinite distance from the diffracting aperture. Fraunhoffer is a special case of Fresnel diffraction.

Single Slit Fraunhoffer Diffraction

In order to find the intensity at point P on the screen as shown in the figure the slit of width 'a' is divided into N parallel strips of width $\Delta$x. Each strip then acts as a radiator of Huygen's wavelets and produces a characteristic wave disturbance at P, whose position on the screen for a particular arrangement of apparatus can be described by the angle $\theta$.

The amplitudes $\Delta$E_o of the wave disturbances at P from the various strips may be taken as equal if $\theta$is not too large.

The intensity is proportional to the square of the amplitude. If I_m represents the intensity at O, its value at P is

${\text{l}}_{\theta }\text{=}{\text{l}}_{\text{m}}{\left(\frac{\sin \alpha }{\alpha }\right)}^2;$ $where\alpha =\frac{\phi }{2}=\frac{\pi a\sin \theta }{\lambda }$

A minimum occurs when, sin $\alpha$ = 0 and $\alpha$ $\ne$ 0, so $\alpha$= n$\pi$, n = 1, 2, 3...

$\Rightarrow \frac{\pi a\sin \theta }{\lambda }=n\pi \Rightarrow a\sin \theta =n\lambda$

Angular width of central maxima of diffraction pattern = 2$\alpha$₁ = 2 sin^‑1($\lambda$/a)

[ $\theta$₁ gives the angular position of first minima]

The concept of diffraction is also useful in deciding the resolving power of optical instruments.

Illustration 1: Light of wavelength 6 x 10^-5cm falls on a screen at a distance of 100 cm from a narrow slit. Find the width of the slit if the first minima lies 1mm on either side of the central maximum.

Solution: Here n = 1, $\lambda$ = 6 x 10^-5 cm.

Distance of screen from slit = 100 cm.

Distance of first minimum from central maxima = 0.1 cm.

sin$\theta$ = $\frac{Dis\tan ceof{1}^{st}\min imafromthecentral\max ima}{Dis\tan ceofthescreenfromtheslit}$

$\theta$₁ = $\frac{0.1}{100}=\frac{1}{1000}$

We know that asin$\theta$ = n$\lambda$

a = $\frac{\lambda }{{\theta }_1}$ = 0.06 cm.

Illustration 2: In YDSE if the source consists of two wavelengths $\lambda$₁ = 4000Å and $\lambda$₂ = 4002Å . Find the distance from the centre where the fringes disappear, if d=1cm ; D=1 m .

Solution: The fringes disappear when the maxima of $\lambda$₁ fall over the minima of $\lambda$₂. That is

$\frac{p}{{\lambda }_1}-\frac{p}{{\lambda }_2}=\frac{1}{2}$ Where p is the optical path difference at that point.

or p = $\frac{{\lambda }_1{\lambda }_2}{2\left({\lambda }_2-{\lambda }_1\right)}$

Here $\lambda$₁ = 4000Å, $\lambda$₂ = 4002Å

$\therefore$ p = 0.04 cm

In YDSE, p = dy/D

$\therefore$ y = $\frac{D}{d}p=\frac{\left(1000\right)}{10}\left(0.4\right)=40mm$