In $Y.D.S.E.,$ the slits are $0.2 \, cm$ apart and are illuminated by yellow light $(\lambda = 600 \, nm).$ What would be the fringe width on a screen placed $1 \, m$ from the plane of the slits if the whole system is immersed in water of refractive index $\mu = \frac{4}{3}$ (in $, mm$)?

Two slits are separated by $0.3\, mm$. $A$ beam of $500\, nm$ light strikes the slits producing an interference pattern. Find the number of maxima observed in the angular range $-30^{\circ} < \theta < 30^{\circ}$.

In Young's double slit experiment,if the wavelength of the light used is doubled and the distance between the two slits is halved,the resultant fringe width becomes $n$ times the initial fringe width. Find the value of $n$.

The width of the fringe is $2\,mm$ on the screen in a double-slit experiment for light of wavelength $400\,nm$. The width of the fringe for light of wavelength $600\,nm$ will be $..............\,mm$.

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