In Young's double-slit experiment,the maximum intensity is $I_0$. If one slit is closed,the new maximum intensity is:

In Young's double-slit experiment,the slits are separated by $0.12 \, mm$ and the screen is at a distance of $1 \, m$. Find the distance of the $3^{rd}$ bright fringe from the center of the screen in $cm$. Given $\lambda = 6000 \, \mathring{A}$.

In an interference experiment,the distance between a maximum and the adjacent minimum is .......

$A$ fringe width of $6 \ mm$ was produced for two slits separated by $1 \ mm$ apart. The screen is placed $10 \ m$ away. The wavelength of light used is $'x' \ nm$. The value of $'x'$ to the nearest integer is

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

If the slit widths of $YDSE$ are in the ratio $1:4$,then the ratio of maximum to minimum intensity on the screen is: (in $:1$)