In $Y.D.S.E.,$ the slits are $0.2 \, cm$ apart and are illuminated by yellow light $(\lambda = 600 \, nm).$ What would be the fringe width on a screen placed $1 \, m$ from the plane of the slits if the whole system is immersed in water of refractive index $\mu = \frac{4}{3}$ (in $, mm$)?

In Young's double-slit experiment,the angular width of the fringes formed on the screen is $\pi / 200$. If light of wavelength $4800 \, \mathring{A}$ is used,find the distance between the slits.

In Young's double-slit experiment using monochromatic light,the shape of the interference pattern on the screen is .....

In a Young's double slit experiment, the slits are separated by $0.3 \, mm$ and the screen is $1.5 \, m$ away from the plane of slits. The distance between the fourth bright fringes on both sides of the central bright fringe is $2.4 \, cm$. The frequency of light used is $.......... \times 10^{14} \, Hz$.

In Young's double slit experiment,the distance between two sources is $0.1 \, mm$. The distance of the screen from the sources is $20 \, cm$. The wavelength of light used is $5460 \, \mathring{A}$. The angular position of the first dark fringe is: (in $^\circ$)

In a Young's double slit experiment, the separation between the two slits is $d$ and the wavelength of the light is $\lambda$. The intensity of light falling on slit $1$ is four times the intensity of light falling on slit $2$. Choose the correct choice(s).
$(A)$ If $d = \lambda$, the screen will contain only one maximum
$(B)$ If $\lambda < d < 2\lambda$, at least one more maximum (besides the central maximum) will be observed on the screen
$(C)$ If the intensity of light falling on slit $1$ is reduced so that it becomes equal to that of slit $2$, the intensities of the observed dark and bright fringes will increase
$(D)$ If the intensity of light falling on slit $2$ is increased so that it becomes equal to that of slit $1$, the intensities of the observed dark and bright fringes will increase