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

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.

The maximum number of possible interference maxima for slit separation equal to $1.8 \lambda$,where $\lambda$ is the wavelength of light used,in a Young's double slit experiment is

In a $YDSE$ experiment,if a slab whose refractive index can be varied is placed in front of one of the slits,then the variation of resultant intensity at the mid-point of the screen with $\mu$ will be best represented by $(\mu \geq 1)$. [Assume slits of equal width and there is no absorption by the slab]

Sodium light $(\lambda = 6 \times 10^{-7} \ m)$ is used to produce an interference pattern. The observed fringe width is $0.12 \ mm$. The angle between the two wave trains is:

In Young's double-slit experiment,we get $10$ fringes in the field of view using monochromatic light of wavelength $4000 \text{ } \mathring{A}$. If we use monochromatic light of wavelength $5000 \text{ } \mathring{A}$,then the number of fringes obtained in the same field of view is