a) Diffraction is the bending or spreading of waves that encounter an object ( a barrier or an opening) in their path.
b) In Fresnel class of diffraction, the source and/or screen are at a finite distance from the aperture.
c) 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 Δx. 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 θ
The amplitudes ΔEo of the wave disturbances at P from
the various strips may be taken as equal if θis not too large.
The intensity is proportional to the square of the amplitude.
If Im represents the intensity at O, its value at P is
The concept of diffraction is also useful in deciding the resolving power of optical instruments.
Illustration 7: Light of wavelength 6 λ 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, λ = 6 ´ 10-5 cm.
Distance of screen from slit = 100 cm.
Distance of first minimum from central maxima = 0.1 cm.
We know that a sinθ = nλ
a = λ/θ1 = 0.06 cm.
Illustration 9:A beam of light consisting of two wavelengths 6500A°and 5200A° is used to obtain interference fringes in a Young’s double slit experiment
(i) Find the distance of the third fringe on the screen from the central maximum for the wavelength 6500A°.
(ii) What is the least distance from the central maximum where the bright fringes due to both wavelength coincide?
(iii) The distance between the slits is 2mm and the distance between the plane of the slits and screen is 120cm. What is the fringe width for l = 6500A°?