Question

A parallel beam of monochromatic light falls normally on a single slit of width \(a\) and a diffraction pattern is observed on a screen placed at distance \(D\) from the slit. Explain:
(I) the formation of maxima and minima in the diffraction pattern, and
(II) why the maxima go on becoming weaker and weaker with increasing order \(n\).

Hint:

For single-slit diffraction: \[ a\sin\theta=n\lambda \] gives the condition for minima. The central maximum is the brightest and has approximately twice the width of each secondary maximum.

Answer 1

Formation of Diffraction Pattern Concept: Diffraction is the bending and spreading of light waves when they pass through a narrow aperture whose dimensions are comparable to the wavelength of light. According to Huygens' principle, every point on the wavefront inside the slit acts as a source of secondary wavelets. The superposition of these wavelets at different points on the screen produces alternate bright and dark regions known as the diffraction pattern. (I) Formation of Maxima and Minima

Step 1: Formation of Central Maximum When light reaches the centre of the screen, all secondary wavelets emerging from different parts of the slit travel equal distances. Hence all wavelets arrive in the same phase and interfere constructively. Therefore a bright fringe of maximum intensity is obtained at the centre and is called the central maximum. \[ \boxed{\theta=0} \] corresponds to the central bright fringe.

Step 2: Formation of Minima Consider light observed at an angle \(\theta\). The path difference between light coming from the top and bottom edges of the slit is \[ a\sin\theta. \] For the first minimum, \[ a\sin\theta=\lambda. \] The slit can be divided into two equal halves. Each point in the upper half has a corresponding point in the lower half whose path difference is \[ \frac{\lambda}{2}. \] These pairs of wavelets interfere destructively and cancel each other completely. Hence intensity becomes zero and a dark fringe is obtained. Therefore minima occur when \[ \boxed{ a\sin\theta=n\lambda } \] where \[ n=1,2,3,\ldots \]

Step 3: Formation of Secondary Maxima Between two successive minima complete cancellation does not occur. A small resultant amplitude survives due to incomplete destructive interference. This produces secondary bright fringes known as secondary maxima. These maxima occur approximately between successive minima. (II) Why Successive Maxima Become Weaker

Step 1: Understand amplitude distribution. As the observation point moves farther from the centre, the phase differences among wavelets emerging from different parts of the slit increase. Consequently a larger fraction of the wavelets cancels one another.

Step 2: Effect on resultant amplitude. The resultant amplitude of successive maxima decreases rapidly with increasing order. Since intensity is proportional to the square of amplitude, \[ I\propto A^2, \] a small decrease in amplitude causes a large decrease in intensity.

Step 3: Conclude the behaviour of maxima. Hence the central maximum is the brightest. The first secondary maximum is much weaker than the central maximum. The second secondary maximum is weaker still, and so on. Therefore the intensity of maxima decreases continuously with increasing order \(n\). \[ \boxed{ I_0>I_1>I_2>I_3>\cdots } \]