Assertion: $A$ white source of light during interference forms only white and black fringes.
Reason: Width of fringe is inversely proportional to the wavelength of the light used.

In Young's double-slit experiment,the slits are separated by $0.12 \, mm$ and the screen is at a distance of $1 \, m$. Find the distance of the $3^{rd}$ bright fringe from the center of the screen in $cm$. Given $\lambda = 6000 \, \mathring{A}$.

In Young's double slit experiment,$62$ fringes are seen in the visible region for sodium light of wavelength $5893 \ \mathring{A}$. If violet light of wavelength $4358 \ \mathring{A}$ is used in place of sodium light,then the number of fringes seen will be:

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.)

In a Young's double slit experiment,the ratio of the amplitude of light coming from the slits is $2:1$. The ratio of the maximum to minimum intensity in the interference pattern is:

Four monochromatic and coherent sources of light,emitting waves in phase of wavelength $\lambda$,are placed at the points $x = 0, d, 2d$,and $3d$ on the $x$-axis. Then: