In Young's double-slit experiment,the width of one slit is twice that of the other. What is the ratio of the maximum to the minimum intensity?

Two coherent sources $P$ and $Q$ produce interference at point $A$ on the screen,where a dark band is formed between the $4^{\text{th}}$ and $5^{\text{th}}$ bright band. The wavelength of light used is $6000 \text{ Å}$. The path difference between $PA$ and $QA$ is:

In Young's double slit experiment,the two slits act as coherent sources of equal amplitude $A$ and wavelength $\lambda$. In another experiment with the same setup,the two slits are of equal amplitude $A$ and wavelength $\lambda$ but are incoherent. The ratio of the intensity of light at the mid-point of the screen in the first case to that in the second case is

In a Young's double-slit experiment,the intensity at a point where the path difference is $\lambda$ is $K$. What will be the intensity at a point where the path difference is $\lambda / 4$?

In a double-slit interference experiment, the fringe width obtained with light of wavelength $5900 \ \mathring{A}$ was $1.2 \ \text{mm}$ for parallel narrow slits placed $2 \ \text{mm}$ apart. In this arrangement, if the slit separation is increased by one-and-a-half times the previous value, then the fringe width is: (in $\text{mm}$)

Young's double-slit experiment is performed with a wavelength of $550 \,nm$. If the distance between the two slits is $1.10 \,mm$ and the distance between the screen and the slits is $1 \,m$,then the distance between two consecutive bright or dark fringes is .....$mm$.