In a Young's double-slit experiment,the separation of the two slits is doubled. To keep the same spacing of fringes,the distance $D$ of the screen from the slits should be made

In Young's double-slit experiment,the position of the $5^{\text{th}}$ bright fringe from the central maximum is $5\,cm$. The distance between the slits and the screen is $1\,m$ and the wavelength of the monochromatic light used is $600\,nm$. The separation between the slits is $........\mu m$.

Two sources of light are $0.6 \, mm$ apart and the screen is placed at a distance of $1.2 \, m$ from them. $A$ light of wavelength $6000 \, \mathring{A}$ is used. The phase difference between the two light waves interfering on the screen at a point at a distance of $3 \, mm$ from the central bright band is:

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).

In Young's double slit experiment,if one of the slits is closed fully,then in the interference pattern:

In Young's double-slit experiment,the intensity of light at a point on the screen where the path difference is $\lambda$ is $k$ units; $\lambda$ being the wavelength of light used. The intensity at a point where the path difference is $\lambda/4$ will be: