Two slits,$4 \, mm$ apart,are illuminated by light of wavelength $6000 \, \mathring{A}$. What will be the fringe width on a screen placed $2 \, m$ from the slits in $mm$?

Light of wavelength $520\, nm$ passing through a double slit produces an interference pattern of relative intensity versus deflection angle $\theta$ as shown in the figure. The separation $d$ between the slits is

Assertion: In Young's double slit experiment,if the wavelength of incident monochromatic light is doubled,the number of bright fringes on the screen will increase.
Reason: The maximum number of bright fringes on the screen is directly proportional to the wavelength of light used.

In an interference pattern,if the ratio of slit widths is $1:9$,find the ratio of maximum to minimum intensity.

$A$ 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?

In a double-slit experiment,the angular width of a fringe is found to be $0.2^{\circ}$ on a screen placed $1 \; m$ away. The wavelength of light used is $600 \; nm$. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water (in $^{\circ}$)? Take the refractive index of water to be $4/3$.