What is effect on the interference fringes in a Young's double slit experiment due to each of the following operations :
(a) the screen is moved away from the plane of the slits,
(b) the monochromatic source is replaced by another monochromatic source of shorter wavelength,
(c ) the separation between the two slits is increased,
(d) the source slit is moved closer to the double slit plane,
(e) the width of the source slit is increased.
(f) the width of two slits are increased,
(g) the monochromatic source is replaced by a source of white light ?
(In each operation, take all parameters, other than the one specified, to remain unchanged) NCERT Solved Example

Answer

(a) Angular separation of the fringes remains constant `(= lambda//d)`. The actual separation of the fringe increases in proportion to the distance of the screen from the plane of the two slits.
(b) The separation of the fringes (and also angular separation) decrease.
(c) The separation of the fringe (and also angular sepration) dereases. (d) By slightly increasing the width of the slits, we are only increasing the intensity of incident beam. Again no change in `lambda,D,d` so `beta` unchaged but sharpness of the fringe increase.

Explanation

We know `beta=(lambdaD)/(d)`
(a) if `D` is increased `rArr beta` increases.
so the width of interference fringe increases.
(b) `lambda` decreases `rArr beta` decrease again.
(c ) `d` is increased `rArr` beta decreases.
If the source slit is moved nearer. no effects on `lambda,D` or `d,` so `beta` remains same.
(d) By slightly increasing the width of the slits, we are only increasing the intensity of incident beam. Again no change in `lambda,D,d.` so `beta` unchanged but sharpness of the fringe increase.