In a Young's double slit experiment,the ratio of the slit widths is $4 : 1$. The ratio of the intensity of maxima to minima,close to the central fringe on the screen,will be

In a Young's double slit experiment with slit separation $0.1\, mm$,one observes a bright fringe at angle $\frac{1}{40}\, rad$ by using light of wavelength $\lambda_1$. When the light of wavelength $\lambda_2$ is used,a bright fringe is seen at the same angle in the same setup. Given that $\lambda_1$ and $\lambda_2$ are in the visible range ($380\, nm$ to $740\, nm$),their values are:

In Young’s double slit experiment,a minimum is obtained when the phase difference of superimposing waves is:

In a two-slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by $5 \times 10^{-2} \, m$ towards the slits, the change in fringe width is $3 \times 10^{-5} \, m$. If the separation between the slits is $10^{-3} \, m$, the wavelength of light used is: (in $\text{Å}$)

In a Young's double-slit experiment,the slits are separated by $0.28 \; mm$ and the screen is placed $1.4 \; m$ away. The distance between the central bright fringe and the fourth bright fringe is measured to be $1.2 \; cm$. Determine the wavelength of light used in the experiment (in $nm$).

In Young's double-slit experiment,if $L$ is the distance between the slits and the screen upon which the interference pattern is observed,$x$ is the average distance between adjacent fringes,and $d$ is the slit separation,then the wavelength of light is given by: