In Young's double slit experiment,when two light waves form the third minimum,they have:

Explain Young's double-slit experiment,its experimental arrangement,and how it produces an interference pattern.

$A$ 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,the slits are $0.5 \, mm$ apart and the interference is observed on a screen at a distance of $100 \, cm$ from the slits. It is found that the $9^{th}$ bright fringe is at a distance of $7.5 \, mm$ from the $2^{nd}$ dark fringe on the same side of the central fringe. The wavelength of the light used is ..... $\mathring{A}$.

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:

Sodium light $(\lambda = 6 \times 10^{-7} \ m)$ is used to produce an interference pattern. The observed fringe width is $0.12 \ mm$. The angle between the two wave trains is: