In Young's double slit experiment,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 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,$I/I_0$ is equal to

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?

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 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:

In Young's double slit experiment,the central bright fringe can be identified