In an interference experiment,the third bright fringe is obtained using light of wavelength $700 \, nm$. What should be the wavelength of light to obtain the fifth bright fringe at the same point?

The distance between two coherent sources is $0.1 \, mm$. The fringe-width on a screen $1.2 \, m$ away from the source is $6.0 \, mm$. The wavelength of light used is ......... $\mathring{A}$.

The intensity of each source in Young's double slit experiment is $I_0$. The distance between the slits is $d = 5\lambda$,where $\lambda$ is the wavelength of the monochromatic light used. What will be the intensity of light in front of one of the slits on a screen (where the slit and screen are at a distance $D = 10d$)?

In Young's double slit experiment, carried out with light of wavelength $\lambda = 5000\;\mathring{A}$, the distance between the slits is $d = 0.2\;mm$ and the screen is at $D = 200\;cm$ from the slits. The central maximum is at $x = 0$. The third maximum (taking the central maximum as the zeroth maximum) will be at $x$ equal to......$cm$.

In Young's double-slit experiment,if the distance between the slits is reduced to half and the distance between the slit and the screen is doubled,then the fringe width

In Young's double slit experiment,the distance between the slits is $1 \,mm$ and the distance between the slit and the screen is $1 \,m$. If the $10^{th}$ fringe is $5 \,mm$ away from the central bright fringe,then the wavelength of the light used will be.....$\mathring A$.