In Young's double slit experiment,the ratio of intensity of maxima and minima in the interference experiment is $25 : 9$. The ratio of the widths of the two slits is:

In Young's double-slit experiment,a monochromatic light source is used to illuminate slits $S_1$ and $S_2$. An interference pattern is obtained on the screen with a fringe width of $w$. If a mica sheet of thickness $t$ and refractive index $\mu$ is placed in front of one of the slits,the new fringe width $w'$ becomes:

In a Young's double-slit experiment,interference is obtained using electron waves. The electrons are produced by an electron gun. To increase the fringe width,one must:

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$: If Young's double slit experiment is performed in an optically denser medium than air,then the consecutive fringes come closer.
Reason $(R)$: The speed of light reduces in an optically denser medium than air while its frequency does not change.
In the light of the above statements,choose the most appropriate answer from the options given below:

In Young's double slit experiment,the distance between the two slits is $0.1 \,mm$,the distance between the slits and the screen is $1 \,m$ and the wavelength of the light used is $600 \,nm$. The intensity at a point on the screen is $75 \%$ of the maximum intensity. Find the smallest distance in $mm$ of this point from the central fringe.

In a Young's double-slit experiment,the fringe width is $0.8 \, mm$ when light of wavelength $6000 \, \mathring{A}$ is used,and the screen is at a distance of $2.5 \, m$. If the experiment is performed in a liquid with a refractive index of $1.6$,the new fringe width will be.....$mm$.