In a Young's double slit experiment,the slit separation is $0.2 \ cm$,the distance between the screen and slit is $1 \ m$. The wavelength of the light used is $5000 \ \mathring{A}$. The distance between two consecutive dark fringes (in $mm$) is

In Young's double slit experiment,when the wavelength used is $6000 \ \mathring{A}$ and the screen is $40 \ cm$ from the slits,the fringes are $0.012 \ cm$ wide. What is the distance between the slits in $cm$?

In a Young's double-slit experiment,two slits act as coherent sources of waves of equal wavelength $\lambda$ and amplitude $A$. In another experiment,the two slits are arranged to act as incoherent sources of waves of the same amplitude and wavelength. If the intensity at the central point of the screen in the first case is $I_1$ and in the second case is $I_2$,then the ratio $I_1/I_2$ 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 $YDSE$,$a=2 \, mm$,$D=2 \, m$,and $\lambda=500 \, nm$. Find the distance of the point on the screen from the central maxima where the intensity becomes $50 \%$ of the central maxima (in $\mu m$).

The Young's double slit experiment is performed in succession using blue light of wavelength $4360 \, \mathring{A}$ and green light of wavelength $5460 \, \mathring{A}$. If the distance of the fourth maximum from the central maximum is $x$,then: