Describe Young's double slit experiment.

Explanation

Expression of fringe width is `beta=(lamda)/(d).....(1)`
(a) From equation (1), `betapropD`
`implies` When screen is moved away from the plane of slits, D will increase and so `beta` wil increase. Hence interference fringes wil become wider.
(b) From equation (1) `betaproplamda`
`implies` When wavelength is made shorter, `lamda` will decrease and so fringe width B will also decrease. Hence interference fringes will become narrower.
(c) From equation (1). `betaprop(1)/(d)`
`implies` When separation between two slits is increased, value of d will increase and so fringe width `beta` will decrease and so interference fringes will become narrower.
When source is moved gradually closer to plane of slits, bright and dark fringes start merging into each other, which causes decrease in the sharpness of fringes. Ultimately, when source is brought very close to the slits, fringes disappear and whole screen is illuminated with average intensity of light
As long as fringes appear on the screen, fringe width `(beta)` remains same because it does not depend on the distance between source and slits.
(e) When width of source is increased, bright and dark fringes start merging which causes decrease in sharpness. Ultimately, at one stage, fringes disappear. Here, also till the fringes appear, fringe width `(beta)` is not affected by change in width of source.
(f) When source of white light is used interference patterns of all component colours overlap on each other with changing phase differences (i.e. incoherently). In this case, central bright fringes for all component colours occur at same central position on the screen, which makes it white in colour (because of mixture of all component colours in equal proportions).
Consider point P on the screen where path difference is `(r_(2)-r_(1))=2000Å=(1)/(2)=(lamda_(b))/(2)` and so there would be destructive interference for blue colour and so blue colour will be absent at point P. Other colours will appear in certain proportion because of partial constructive interferences.
Consider another point Q on the screen such that path difference at this point is `(r_(2)-r_(1))=4000Å=(1)/(2)(8000Å)=(lamda_(b))/(2)` and so there would be destructive interference for red colour and so red colour will be absent at point Q but blue colour will be present at this point with maximum intensity because of constructive interference. For remaining colours, there would be partial constructive (or partial destructive interferences) at point Q. Hence blue colour would be dominant at point Q.
Thus, the fringes closest to central white fringe, will appear red and the fringes farthest from central white fringe will appear blue. After few fringes, no clear fringes are seen on the screen.