In an interference experiment,the spacing between successive maxima or minima is (where the symbols have their usual meanings):

In a Young's double slit experiment, the slits are separated by $0.3 \, mm$ and the screen is $1.5 \, m$ away from the plane of slits. The distance between the fourth bright fringes on both sides of the central bright fringe is $2.4 \, cm$. The frequency of light used is $.......... \times 10^{14} \, Hz$.

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 a Young's double-slit experiment,the experiment is performed with blue light of wavelength $4360 \; \mathring{A}$ and green light of wavelength $5460 \; \mathring{A}$. If the distance of the $4^{th}$ bright fringe from the central maximum is $x$,then:

In Young's double-slit experiment,the slits are separated by $0.2 \, cm$ and illuminated by light of wavelength $\lambda = 5896 \, \mathring{A}$. What is the fringe width on a screen placed $1 \, m$ away from the plane of the slits? What will be the fringe width if the entire system is immersed in water (in $, mm$)? (Refractive index of water $\mu_w = 4/3$)

In Young's experiment,the ratio of maximum to minimum intensities of the fringe system is $4:1$. The amplitudes of the coherent sources are in the ratio: (in $:1$)