The maximum number of possible interference maxima for slit separation equal to $1.8 \lambda$,where $\lambda$ is the wavelength of light used,in a Young's double slit experiment is

The intensity of each source in Young's double slit experiment is $I_0$. The distance between the slits is $d = 5\lambda$,where $\lambda$ is the wavelength of the monochromatic light used. What will be the intensity of light in front of one of the slits on a screen (where the slit and screen are at a distance $D = 10d$)?

In a Young's double slit experiment,the intensity at a point where the path difference is $\frac{\lambda}{6}$ ($\lambda$ being the wavelength of the light used) is $I$. If $I_0$ denotes the maximum intensity,$I/I_0$ is equal to

$A$ beam of monoenergetic electrons,which have been accelerated from rest by a potential $U$,is used to form an interference pattern in a Young's double slit experiment. The electrons are now accelerated by potential $4U$. Then,the fringe width

In Young's double slit experiment,the distance between the two slits is $0.1 \,mm$,the distance between the slits and the screen is $1 \,m$ and the wavelength of the light used is $600 \,nm$. The intensity at a point on the screen is $75 \%$ of the maximum intensity. Find the smallest distance in $mm$ of this point from the central fringe.

The angular width of a fringe in a double slit experiment is found to be $0.2^{\circ}$ on a screen $1 \ m$ away. The wavelength of light used is $600 \ nm$. What is the change in the angular width of the fringe if the entire measurement system is immersed in water (in $^{\circ}$)? (Use refractive index of water as $\mu = \frac{4}{3}$)