At two points $P$ and $Q$ on the screen in Young's double-slit experiment,waves from slits $S_1$ and $S_2$ have a path difference of $0$ and $\frac{\lambda}{4}$ respectively. The ratio of intensities at $P$ and $Q$ will be

In $YDSE$,let $S_1$ and $S_2$ be the two slits,and $C$ be the centre of the screen. If $\theta$ is the angle $S_1CS_2$ and $\lambda$ is the wavelength,the fringe width will be

$A$ beam of light consisting of two wavelengths $7000 \; \mathring{A}$ and $5500 \; \mathring{A}$ is used to obtain an interference pattern in Young's double-slit experiment. The distance between the slits is $2.5 \; mm$ and the distance between the plane of the slits and the screen is $150 \; cm$. The least distance from the central fringe,where the bright fringes due to both the wavelengths coincide,is $n \times 10^{-5} \; m$. The value of $n$ is .............

In a Young's double-slit experiment, the distance between two coherent sources is $0.90 \, mm$ and the distance from the sources to the screen is $1 \, m$. If the distance of the second bright fringe from the center of the central bright fringe is $1 \, mm$, find the wavelength of the light used.

Four monochromatic and coherent sources of light,emitting waves in phase of wavelength $\lambda$,are placed at the points $x = 0, d, 2d$,and $3d$ on the $x$-axis. Then:

In a double-slit experiment,the angular width of a fringe is found to be $0.2^{\circ}$ on a screen placed $1 \; m$ away. The wavelength of light used is $600 \; nm$. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water (in $^{\circ}$)? Take the refractive index of water to be $4/3$.