In a Young's double slit experiment,the path difference,at a certain point on the screen,between two interfering waves is $1/8^{th}$ of the wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to:

What happens to the fringe width in Young's double-slit experiment if it is performed in glycerine instead of air?

In Young's double-slit experiment,if $L$ is the distance between the slits and the screen upon which the interference pattern is observed,$x$ is the average distance between adjacent fringes,and $d$ is the slit separation,then the wavelength of light is given by:

In $YDSE$ setup,light of wavelength $640 \, nm$ is used with $d = 0.8 \, mm$ and $D = 1 \, m$. If intensity at central maximum is $I_0$ and its position is $y = 0$,then:

Two coherent sources $O_1$ and $O_2$ in Young's double slit experiment are illuminated with monochromatic light of wavelength $5000 \text{ Å}$. If a second order dark fringe is formed at a point $R$ on the screen,the path difference $O_1 R \sim O_2 R$ is (in $\mu m$)

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