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 Young's double slit experiment,the intensity at a point is $1/4$ of the maximum intensity. The angular position of this point is:

The fringe width in a $YDSE$ pattern is $2.4 \times 10^{-4} \, m$ when red light of wavelength $6400 \, \mathring{A}$ is used. How much will it change if blue light of wavelength $4000 \, \mathring{A}$ is used?

In a Young's double slit experiment, the slits are separated by $0.2 \text{ mm}$ and the screen is placed $2.0 \text{ m}$ away. The distance between the central bright fringe and the third bright fringe is measured to be $1.5 \text{ cm}$. Determine the wavelength of light used in the experiment.

In a Young's double-slit experiment,$D$ is the distance of the screen from the slits and $d$ is the separation between the slits. The distance of the nearest point to the central maximum where the intensity is the same as that due to a single slit is equal to:

In the ideal double-slit experiment,when a glass plate (refractive index $\mu = 1.5$) of thickness $t$ is introduced in the path of one of the interfering beams (wavelength $\lambda$),the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is: