In $Y.D.S.E$, two slits are made $1 \, mm$ apart and the screen is placed $1 \, m$ away. The fringe separation when light of wavelength $500 \, nm$ is used is:

In a double-slit experiment using light of wavelength $600 \; nm$,the angular width of a fringe formed on a distant screen is $0.1^{\circ}$. What is the spacing between the two slits?

The source that illuminates the double-slit in a 'double-slit interference experiment' emits two distinct monochromatic waves of wavelengths $\lambda_1 = 500\,nm$ and $\lambda_2 = 600\,nm$. Each wavelength produces its own interference pattern on the screen. At the central point,where the path difference is zero,the maxima of both patterns coincide. As one moves away from the central region,the two fringe systems gradually go out of step. The combined fringe system becomes completely indistinct when a maximum of one wavelength coincides with a minimum of the other. This happens when the path difference in $nm$ is:

Two slits at a distance of $1\, mm$ are illuminated by light of wavelength $6.5 \times 10^{-7}\, m$. The interference fringes are observed on a screen placed at a distance of $1\, m$. The distance between the third dark fringe and the fifth bright fringe will be......$mm$.

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:

In a double slit experiment,when the distance between slits is increased $10$ times,while their distance from the screen is halved,then the fringe width