Obtain an expression for width of central maxima by diffraction of light at a single slit.

The phenomenon of bending of light round the sharp corners and spreading into the regions of the geometrical shadow is called diffraction. The light waves are diffracted only when the size of the obstacle is comparable to the wavelength of the light. In principle, all the types of wave motion exhibit diffraction effect. The diffraction of radio waves and sound waves is readily observed in the daily life. Consider that a monochromatic source of light S , emitting light waves of wavelength λ , is placed at the principal focus of the convex lens L 1 . A parallel beam of light i.e. a plane wavefront gets incident on a narrow slit A B of width a as shown above. The diffraction pattern is obtained on a screen lying at a distance D from the slit and at the focal plane of the convex lens L 2 . According to rectilinear propagation of light, a bright image of the slit is expected at the centre O of the screen. But in practice, we get a diffraction pattern i.e. a central maximum at the centre O flanked by a number of dark and bright fringes called secondary maxima and minima on either side of the point O . The central maximum extends up to the distance y 1 (the distance of first secondary minimum) on both sides of the centre of the screen. Therefore, the width of the central maximum, β o = 2 y 1 But we know that, y n = n λ D a . Thus for n = 1 , β o = 2 λ D a

Derive the expression for the fringe width in Young's double slit experiment.

Consider two coherent sources of light S 1 and S 2 are placed at a distance d apart and a screen is placed at a distance D from the two sources. Let P be a point on the screen at a distance y from the centre of the screen. If x is the path difference between the light waves then, x = S 2 P - S 1 P In right-angled ∆ S 2 B P , we have S 2 P 2 = S 2 B 2 + B P 2 = D 2 + y + d 2 2 Also in right-angled ∆ S 1 A P , we have S 1 P 2 = S 1 A 2 + A P 2 = D 2 + y - d 2 2 Thus, S 2 P 2 - S 1 P 2 = D 2 + y + d 2 2 - D 2 + y - d 2 2 or ( S 2 P + S 1 P ) ( S 2 P - S 1 P ) = y + d 2 2 - y - d 2 2 Since, S 2 P - S 1 P = x (the path difference) Thus, ( S 2 P + S 1 P ) · x = 4 y · d 2 = 2 y d or x = 2 y d S 2 P + S 1 P In practice, S 2 P ≈ S 1 P ≈ D Hence, x = 2 y d 2 D or x = y d D . This is the expression for path difference. The point P will be a centre of bright fringe, if y d D = n λ or y = n λ D d , where, n = 0 , 1 , 2 , 3 , . . . . . . . Similarly, the point P will be a centre of dark fringe, if y d D = ( 2 n + 1 ) λ 2 or y = ( 2 n + 1 ) λ D 2 d , where, n = 0 , 1 , 2 , 3 , . . . . . . . The spacing between any two consecutive bright fringes or dark fringes is called fringe width. It is represented by β . Thus, β = y n - y n - 1 = n λ D d - ( n - 1 ) λ D d = λ D d (for bright fringe) and β ' = y n ' - y n - 1 ' = ( 2 n + 1 ) λ D 2 d - 2 n - 1 + 1 · λ D 2 d = λ D d (for dark fringe) Thus, it is clear that, β = β '