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,the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is

Young's double slit experiment is first performed in air and then in a medium other than air. It is found that the $8^{th}$ bright fringe in the medium lies where the $5^{th}$ dark fringe lies in air. The refractive index of the medium is nearly:

In the Young's double slit experiment,the distance between the slits varies in time as $d(t) = d_{0} + a_{0} \sin \omega t$; where $d_{0}$,$\omega$,and $a_{0}$ are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as:

In Young's double slit experiment,light from two identical sources is superimposing on a screen. The path difference between the two light waves reaching a point on the screen is $\frac{7 \lambda}{4}$. The ratio of the intensity of the fringe at this point with respect to the maximum intensity of the fringe is:

In an interference arrangement similar to Young's double slit experiment,the slits $S_1$ and $S_2$ are illuminated with coherent microwave sources each of frequency $10^6 \ Hz$. The sources are synchronized to have zero phase difference. The slits are separated by distance $d = 150 \ m$. The intensity $I(\theta)$ is measured as a function of $\theta$,where $\theta$ is defined as shown. If $I_0$ is the maximum intensity,then $I(\theta)$ for $0 \le \theta \le 90^\circ$ is given by: