In a Young's double slit experiment,$12$ fringes are observed to be formed in a certain segment of the screen when light of wavelength $600 \ nm$ is used. If the wavelength of light is changed to $400 \ nm$,the number of fringes observed in the same segment of the screen is:

$A$ beam of light consisting of two wavelengths $7000 \; \mathring{A}$ and $5500 \; \mathring{A}$ is used to obtain an interference pattern in Young's double-slit experiment. The distance between the slits is $2.5 \; mm$ and the distance between the plane of the slits and the screen is $150 \; cm$. The least distance from the central fringe,where the bright fringes due to both the wavelengths coincide,is $n \times 10^{-5} \; m$. The value of $n$ is .............

In Young's double-slit experiment, the fringe width is $\beta$. If the entire arrangement is placed in a liquid of refractive index $n$, what does the fringe width become?

In Young's double-slit experiment,the intensity at a point on the screen where the path difference is $\lambda/6$ is $I$. If the intensity of the central bright fringe is $I_0$,then $I/I_0$ is:

In Young's experiment,the distance between slits is $0.28 \,mm$ and the distance between the slits and the screen is $1.4 \,m$. The distance between the central bright fringe and the third bright fringe is $0.9 \,cm$. What is the wavelength of the light used in $\mathring{A}$?

In a double slit experiment,at a certain point on the screen,the path difference between the two interfering waves is $\frac{1}{8}$ of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is: