What is the 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 monochromatic source is replaced by a source of white light?
(In each operation,take all parameters,other than the one specified,to remain unchanged.)

(N/A) The angular separation of the fringes remains constant $(=\lambda / d)$. The actual fringe width $\beta = \lambda D / d$ increases in proportion to the distance $D$ of the screen from the plane of the slits.
$(b)$ The fringe width $\beta = \lambda D / d$ decreases because the wavelength $\lambda$ is smaller.
$(c)$ The fringe width $\beta = \lambda D / d$ decreases because the slit separation $d$ is increased.
$(d)$ Let $s$ be the size of the source and $S$ be its distance from the plane of the two slits. For interference fringes to be visible,the condition $s / S < \lambda / d$ must be satisfied. As $S$ decreases,the condition becomes harder to satisfy. The interference pattern becomes less sharp,and eventually,the fringes disappear.
$(e)$ Similar to $(d)$,as the source slit width $s$ increases,the condition $s / S < \lambda / d$ is violated. The interference pattern becomes less sharp and eventually disappears.
$(f)$ The interference patterns for different wavelengths overlap. The central bright fringe is white. Since fringe width $\beta \propto \lambda$,the fringes for shorter wavelengths (blue) are closer to the center than those for longer wavelengths (red). Thus,the fringes appear coloured,with red on the outside and blue on the inside,eventually becoming white/blurred.