If the distance between the first maxima and the fifth minima of a double slit pattern is $7\,mm$ and the slits are separated by $0.15\,mm$ with the screen $50\,cm$ from the slits,then find the wavelength of the light used in $nm$.

In a Young's double slit experiment,the width of one of the slits is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

In Young's double slit experiment,$62$ fringes are seen in the visible region for sodium light of wavelength $5893 \ \mathring{A}$. If violet light of wavelength $4358 \ \mathring{A}$ is used in place of sodium light,then the number of fringes seen will be:

In Young's double slit experiment,the slits are $0.5 \, mm$ apart and the interference is observed on a screen at a distance of $100 \, cm$ from the slits. It is found that the $9^{th}$ bright fringe is at a distance of $7.5 \, mm$ from the $2^{nd}$ dark fringe on the same side of the central fringe. The wavelength of the light used is ..... $\mathring{A}$.

In Young's double slit experiment,the amplitudes of two sources are $3a$ and $a$ respectively. The ratio of intensities of bright and dark fringes will be (in $:1$)

In a Young's double-slit experiment,the wavelength of light used is $5000 \, \mathring{A}$ and the fringe width obtained is $1 \, mm$. If the wavelength of light is changed to $6000 \, \mathring{A}$ while keeping the experimental setup unchanged,the new fringe width will be ........ $mm$.