The distance between two slits is $3 \ mm$ and the screen is placed at $2 \ m$ distance. When blue-green light of wavelength $500 \ nm$ is used, then the distance between two consecutive fringes will be (in $mm$)?

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

In the ideal double-slit experiment,when a glass plate (refractive index $\mu = 1.5$) of thickness $t$ is introduced in the path of one of the interfering beams (wavelength $\lambda$),the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is:

In a Young's double slit experiment,the distance between the two identical slits is $6.1$ times larger than the slit width. Then the number of intensity maxima observed within the central maximum of the single slit diffraction pattern is

In a Young's double-slit experiment,the fringe pattern is observed on a screen placed at a distance $D$. The slits are separated by $d$ and are illuminated by light of wavelength $\lambda$. Find the distance from the central point where the intensity falls to half the maximum intensity.

In a double slit experiment,at a certain point on the screen,the path difference between the two interfering waves is $\frac{1}{8}$ of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is: