Young's double-slit experiment is performed using microwaves of wavelength $\lambda = 3 \, cm$. The distance between the plane of the slits and the screen is $D = 100 \, cm$,and the distance between the slits is $d = 5 \, cm$. Find $(a)$ the number of maxima and $(b)$ their positions on the screen.

$A$ beam of light consisting of two wavelengths, $650\; nm$ and $520\; nm$, is used to obtain interference fringes in a Young's double-slit experiment.
$(a)$ Find the distance of the third bright fringe on the screen from the central maximum for wavelength $650\; nm$.
$(b)$ What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

$A$ Young's double slit interference arrangement with slits $S_1$ and $S_2$ is immersed in water (refractive index $\mu_w = 4/3$) as shown in the figure. The positions of maxima on the surface of water are given by $x^2 = p^2 m^2 \lambda^2 - d^2$,where $\lambda$ is the wavelength of light in air (refractive index $\mu_a = 1$),$2d$ is the separation between the slits,and $m$ is an integer. The value of $p$ is

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.

The correct curve between fringe width $\beta$ and the distance between the slits $(d)$ is:

$A$ double slit of separation $1.5\, mm$ is illuminated by white light (between $4500\, \mathring{A} - 7000\, \mathring{A}$). On a screen $120\, cm$ away,a coloured interference pattern is formed. If a pinhole is made on this screen at a distance $3.0\, mm$ from the central white fringe,which of the following wavelengths will be absent in the transmitted light? (in $\mathring{A}$)