The ratio of the intensity at the centre of a bright fringe to the intensity at a point one-quarter of the fringe width from the centre is

In Young's double slit experiment,monochromatic light of wavelength $5000 \ \mathring{A}$ is used. The slits are $1.0 \ mm$ apart and the screen is placed at $1.0 \ m$ away from the slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is . . . . . . $\times 10^{-6} \ m$.

In the Young's double slit experiment,the intensity of light at a point on the screen where the path difference is $\lambda$ is $K$ ($\lambda$ being the wavelength of light used). The intensity at a point where the path difference is $\lambda / 4$ will be:

In a double slit experiment,the separation between the slits is $d = 0.25 \, cm$ and the distance of the screen $D = 100 \, cm$ from the slits. If the wavelength of light used is $\lambda = 6000 \mathring{A}$ and $I_0$ is the intensity of the central bright fringe,the intensity at a distance $x = 4 \times 10^{-5} \, m$ from the central maximum is

If the intensity of each wave in the observed interference pattern in Young's double slit experiment is $I_0$,then for some point $P$ where the phase difference is $\phi$,the resultant intensity $I$ will be:

In a Young's double slit experiment,the intensity at a point where the path difference is $\frac{\lambda}{6}$ ($\lambda$ being the wavelength of the light used) is $I$. If $I_0$ denotes the maximum intensity,the ratio $\frac{I_0}{I}$ is equal to: