In Young's double slit experiment,the $y$-coordinates of the central maxima and the $10^{th}$ maxima are $2 \, cm$ and $5 \, cm$ respectively. When the $YDSE$ apparatus is immersed in a liquid of refractive index $\mu = 1.5$,what will be the corresponding $y$-coordinates?

The intensity of each source in Young's double slit experiment is $I_0$. The distance between the slits is $d = 5\lambda$,where $\lambda$ is the wavelength of the monochromatic light used. What will be the intensity of light in front of one of the slits on a screen (where the slit and screen are at a distance $D = 10d$)?

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 a double slit experiment,the distance between slits is increased $10$ times,whereas their distance from the screen is halved. The fringe width:

In Young's double slit experiment,the seventh maximum with wavelength $\lambda_1$ is at a distance $d_1$ from the central maximum,and the same maximum with wavelength $\lambda_2$ is at a distance $d_2$. Then $d_1/d_2$ is:

Light of wavelength $600 \, nm$ is incident on a double slit and the interference fringes are formed on a screen $1 \, m$ apart. The separation between two consecutive dark fringes on a screen is found to be $1.2 \, mm$. What is the separation between the slits in $mm$?