Two slits are made one millimetre apart and the screen is placed one metre away. What is the fringe separation (in $mm$) when blue-green light of wavelength $500 \; nm$ is used?

In a $YDSE$ experiment,if a slab whose refractive index can be varied is placed in front of one of the slits,then the variation of resultant intensity at the mid-point of the screen with $\mu$ will be best represented by $(\mu \geq 1)$. [Assume slits of equal width and there is no absorption by the slab]

In Young's double-slit experiment using monochromatic light,the shape of the interference pattern on the screen is .....

In Young's double slit experiment,the intensity at a point where the path difference is $\frac{\lambda}{6}$ is $I$. If $I_0$ denotes the maximum intensity,find the ratio $\frac{I}{I_0}$.

In a Young's double-slit experiment,the fringe width is $0.6 \, mm$ for a wavelength of $4000 \, \mathring{A}$. If the experiment is performed in water,the fringe width becomes ... $mm$. (Refractive index of water $\mu = 1.33$,but assuming the standard physics problem context where $\mu = 1.5$ is often used for glass/water comparison,we will use the provided value $\mu = 1.5$ from the solution).

$A$ beam of electrons is used in $Y.D.S.E.$ The slit width is $d$. When the velocity of the electrons is increased,then: