Question:

In diffraction experimentation monochromatic light which passes through a single slit illuminates a screen. A pattern of fringes seems on the screen. The pattern comprises a central bright fringe. What occurs to the width of the central? bright fringe if the slit is substitute by a more narrow slit?

Answer:

First we have to keep in mind that monochromatic light is light of a single wavelength. After that that in your physics course we use the symbol w to represent the width of a single slit. The equation ml = wsinq, (with m being 1, 2, 3, ...) relates the angle at which minima take place in the case of single-slit diffraction to the wavelength l of the light and the width of the slit w.

The central diffraction maximum expands from the minimum that is just to the left of centre all the way to the minimum that is just to the right of centre. Therefore the width of the central diffraction maximum is the distance between the two minima. Setting m = 1 in the expression for the diffraction minima permits us to get the angular separation between the straight-ahead direction and the direction for either first minimum. Therefore for any one slit-to-screen distance the greater the angle q for the case m = 1 the greater the width of the central maximum. Setting m = 1 in the diffraction equation ml = wsinq as well as solving for sinq we find sinq = l/w. Because w is in the denominator the smaller w is the bigger sinq is. As well as for the range of angles under consideration (0 to 90 degrees) the bigger sinq is the bigger q itself is and hence the wider the central maximum.