In Young's double slit experiment,the distance between the sources is $1 \ mm$ and the distance between the screen and the source is $1 \ m$. If the fringe width on the screen is $0.06 \ cm$,then $\lambda = \dots \mathring{A}$.

In a double-slit experiment,the angular width of a fringe is found to be $0.2^{\circ}$ on a screen placed $1 \; m$ away. The wavelength of light used is $600 \; nm$. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water (in $^{\circ}$)? Take the refractive index of water to be $4/3$.

In $Y.D.S.E$,the ratio of intensity of maxima and minima is $25 : 9$. The ratio of slit width is

The path difference between two interfering light waves meeting at a point on the screen is $\left(\frac{87}{2}\right) \lambda$. The band obtained at that point is

In $Y.D.S.E.$ using red and blue lights of wavelengths $7800 \, \mathring{A}$ and $5200 \, \mathring{A}$,the $n^{th}$ red fringe coincides with the $(n + 1)^{th}$ blue fringe. The value of $n$ is:

The ratio of intensities at two points on the screen in Young's double slit experiment when waves from the two slits have a path difference of $0$ and $\lambda/4$ ($\lambda$ is the wavelength of light used) $(\cos 0^\circ = 1, \cos \pi/2 = 0)$.