Question : What is the effect on the interference fringes in a Young's double slit experiment when the monochromatic source is replaced by a source of white light?
Doubt by Aishwarya
Solution :
Short Answer : The central fringe is white. The fringe closest on either side of the central white fringe is red and the farthest will appear blue.
Long Answer :
We know, In Young's Double Slit Experiment
Fringe Width (ꞵ) = λD/d
xn = nλD/d
xn'=(2n+1)λD/2d
Fringe Width (ꞵ) = λD/d
xn = nλD/d
xn'=(2n+1)λD/2d
λRed>λviolet
If monochromatic source of light is replaced by a source of white light, then the interference patterns due to difference constituent colours of white light will overlap. The central bright fringes for all colours are at same point, hence the central fringe is white but all other fringes will be coloured. As we know the fringe width (ꞵ) for violet colour light will be least and that of red light will be maximum. As we move away from the centre of the screen then the first minima for violet light will come first but the fringe will appear red there. Slightly farther away where there is first minima for red light, the fringe will appear violet.
Thus, fringe closest on either side of central white fringe is red and the farthest is violet. After a few coloured fringes, no clear fringe pattern is seen due to overlapping of fringe patterns of different colours.
Thus, fringe closest on either side of central white fringe is red and the farthest is violet. After a few coloured fringes, no clear fringe pattern is seen due to overlapping of fringe patterns of different colours.
Form NCERT
The interference patterns due to different component colours of white light overlap (incoherently). The central bright fringes for different colours are at the same position. Therefore, the central fringe is white. For a point P for which S2P –S1P = λb/2, where λb (≈ 4000 Å) represents the wavelength for the blue colour, the blue component will be absent and the fringe will appear red in colour. Slightly farther away where S2Q–S1Q = λb = λr/2 where λr (≈ 8000 Å) is the wavelength for the red colour, the fringe will be predominantly blue. Thus, the fringe closest on either side of the central white fringe is red and the farthest will appear blue. After a few fringes, no clear fringe pattern is seen.
The interference patterns due to different component colours of white light overlap (incoherently). The central bright fringes for different colours are at the same position. Therefore, the central fringe is white. For a point P for which S2P –S1P = λb/2, where λb (≈ 4000 Å) represents the wavelength for the blue colour, the blue component will be absent and the fringe will appear red in colour. Slightly farther away where S2Q–S1Q = λb = λr/2 where λr (≈ 8000 Å) is the wavelength for the red colour, the fringe will be predominantly blue. Thus, the fringe closest on either side of the central white fringe is red and the farthest will appear blue. After a few fringes, no clear fringe pattern is seen.
