JEE Main & Advanced Physics Wave Optics / तरंग प्रकाशिकी Diffraction at Single Slit (Fraunhoffer Diffraction)

Suppose a plane wave front is incident on a slit AB (of width b). Each and every part of the expose part of the plane wave front (i.e. every part of the slit) acts as a source of secondary wavelets spreading in all directions. The diffraction is obtained on a screen placed at a large distance. (In practice, this condition is achieved by placing the screen at the focal plane of a converging lens placed just after the slit).

(1) The diffraction pattern consists of a central bright fringe (central maxima) surrounded by dark and bright lines (called secondary minima and maxima).

(2) At point O on the screen, the central maxima is obtained. The wavelets originating from points A and B meets in the same phase at this point, hence at O, intensity is maximum.

(3) Secondary minima : For obtaining nth secondary minima at P on the screen, path difference between the diffracted waves \[\Delta =b\sin \theta =n\lambda \]

(i) Angular position of nth secondary minima \[\sin \theta \approx \theta =\frac{n\lambda }{b}\]

(ii) Distance of nth secondary minima from central maxima

\[{{x}_{n}}=D.\theta =\frac{n\lambda D}{b}=\frac{n\lambda f}{b}\]; where D = Distance between slit and screen. \[f\approx D=\] Focal length of converging lens.

(4) Secondary maxima : For nth secondary maxima at P on the screen.

Path difference \[\Delta =b\sin \theta =(2n+1)\frac{\lambda }{2}\]; where n = 1, 2, 3 .....

(i) Angular position of nth secondary maxima

\[\sin \approx \theta \approx \frac{(2n+1)\lambda }{2b}\]

(ii) Distance of nth secondary maxima from central maxima

\[{{x}_{n}}=D.\theta =\frac{(2n+1)\lambda D}{2b}=\frac{(2n+1)\lambda f}{2b}\]

(5) Central maxima : The central maxima lies between the first minima on both sides.

(i) The Angular width d central maxima = \[2\theta =\frac{2\lambda }{b}\]

(ii) Linear width of central maxima \[=2x=2D\theta =2f\theta =\frac{2\lambda f}{b}\]

(6) Intensity distribution : If the intensity of the central maxima is \[{{l}_{0}}\] then the intensity of the first and second secondary maxima are found to be \[\frac{{{I}_{0}}}{22}\] and \[\frac{{{I}_{0}}}{61}\]. Thus diffraction fringes are of unequal width and unequal intensities.

(i) The mathematical expression for in intensity distribution on the screen is given by

\[l={{I}_{o}}{{\left( \frac{\sin \alpha }{\alpha } \right)}^{2}}\]  where \[\alpha \] is just a convenient connection between the angle \[\theta \] that locates a point on the viewing screening and light intensity I.

\[\phi =\] Phase difference between the top and bottom ray from the slit width b.

Also \[\alpha =\frac{1}{2}\varphi =\frac{\pi b}{\lambda }\sin \theta \]

(ii) As the slit width increases (relative to wavelength) the width of the control diffraction maxima decreases; that is, the light undergoes less flaring by the slit. The secondary maxima also decreases in width (and becomes weaker).

(iii) If \[b>>\lambda \], the secondary maxima due to the slit disappear; we then no longer have single slit diffraction.

(iv) When the slit width is reduced by a factor of 2, the amplitude of the wave at the centre of the screen is reduced by a factor of 2, so the intensity at the centre is reduced by a factor of 4.