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

In a Young's double-slit experiment,the intensity at a point where the path difference is $\lambda$ is $K$. What will be the intensity at a point where the path difference is $\lambda / 4$?

In Young's double slit experiment, the angular width of a fringe is found to be $0.2^\circ$ on a screen placed $1 \text{ m}$ away. The wavelength of light used is $600 \text{ nm}$. If the entire apparatus is immersed in water of refractive index $4/3$, the angular width of the fringe will be: (in $^\circ$)

In Young's double-slit experiment,for a distance of $1.8 \lambda$ between the two slits,what is the maximum number of possible interference maxima? Here,$\lambda$ is the wavelength of light.

In Young's double slit experiment,the distance between slits and the screen is $1.0\,m$ and monochromatic light of $600\,nm$ is being used. $A$ person standing near the slits is looking at the fringe pattern. When the separation between the slits is varied,the interference pattern disappears for a particular distance $d_0$ between the slits. If the angular resolution of the eye is $\frac{1}{60}^o,$ the value of $d_0$ is close to......$mm$

In Young's double-slit experiment,$\frac{d}{D} = 10^{-4}$. The intensity at point $P$ on the screen is equal to the intensity of one of the sources. If the wavelength of the light used is $\lambda = 6000 \, \mathring{A}$,what is the distance of point $P$ from the central bright fringe in $mm$?