What is the effect on the interference fringes in a Young' double slit experiment due to each of the following operations.
a. the screen is moved away from the plane of the slits.
b. the (monolchromatic) 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 monochromatic source is replaced by a source of white light?
(in each operation, take all parameters, other than the one specified, to remain unchanged).

Explanation

a. Angular separation of the fringes remains constant `(=(lamda)/d)`. The actual separation of the fringes 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) decreases. See the condition mentioned in (d) below .
c. The separation of the fringes ( and also angular separation) decreases. See, the condition mentioned in d. below.
d. Let s be the size of the source and S its distance from the plane of the two slits.For interference fringes to be seen,the conditioin `s/Slt(lamda)/d` should be satisfied, otherwise, interference patterns produced by different parts of the source overlap and no fringes are seen. Thus,as S decreases(i.e.the source slit is brought closer), the interference pattern gets less and less sharp, and when the source is brought too close for this condition to be valid, the fringes disappear. Till this happens, the fringe separation remains fixed.
e.Same as in d.As the source slit width increases,fringe pattern gets less and less sharp.
When the source slit is so wide that the condition`s/Slt(lamda)/d` is not satisfied, the interference pattern disappears.
f.The interference patterns due to different component colours of white light overlap (incoherently). The central bright fringes for different colours are the same position. Therefore, the central fringe is white. For a point P for which `S_(2)P-S_(1)P=(lamda_(b))/2` where `lamda_(b) (~~4000Å)` 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 `S_(2)Q-S_(1)Q=lamda_(b)=(lamda_(r))/2` where`lamda_(r)(~~8000Å)` 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.