In Young's double-slit experiment,the intensity of light at a point on the screen where the path difference is $\lambda$ is $k$ units; $\lambda$ being the wavelength of light used. The intensity at a point where the path difference is $\lambda/4$ will be:

Laser light $(630 \, nm)$ incident on a pair of slits produces an interference pattern in which the bright fringes are separated by $8.3 \, mm$. $A$ second light produces an interference pattern in which the bright fringes are separated by $7.6 \, mm$. What is the wavelength of this second light in $nm$?

Two coherent sources $P$ and $Q$ produce interference at point $A$ on the screen,where a dark band is formed between the $4^{\text{th}}$ bright band and the $5^{\text{th}}$ bright band. The wavelength of the light used is $6000 \text{ Å}$. The path difference between $PA$ and $QA$ is:

$A$ 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

In Young's double slit experiment, two slits are illuminated with light of wavelength $\lambda$. The line joining $A_1P$ is perpendicular to $A_1A_2$. If the first minimum is detected at $P$, the value of slit separation '$a$' will be (where $D$ is the distance between the source and the screen):

In a Young's double-slit experiment,interference is obtained using electron waves. The electrons are produced by an electron gun. To increase the fringe width,one must: