In Young's double slit experiment,the distance between the slits is $1 \,mm$ and the distance between the slit and the screen is $1 \,m$. If the $10^{th}$ fringe is $5 \,mm$ away from the central bright fringe,then the wavelength of the light used will be.....$\mathring A$.

The distance between two coherent sources is $1 \ mm$. The screen is placed at a distance of $1 \ m$ from the sources. If the distance of the third bright fringe is $1.2 \ mm$ from the central fringe, the wavelength of light used is:

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:

In Young's double slit experiment,light from two identical sources is superimposing on a screen. The path difference between the two light waves reaching a point on the screen is $\frac{7 \lambda}{4}$. The ratio of the intensity of the fringe at this point with respect to the maximum intensity of the fringe is:

In Young's double-slit experiment,if the distance between the slits is twice the wavelength,the number of possible interference maxima is:

In a double slit interference experiment,the fringe width obtained with a light of wavelength $5900 \text{ Å}$ was $1.2 \text{ mm}$ for parallel narrow slits placed $2 \text{ mm}$ apart. In this arrangement,if the slit separation is increased by one-and-half times the previous value,then the fringe width is (in $ \text{ mm}$)