Two slits are made $1\, mm$ apart and the screen is placed $1\, m$ away. The fringe separation when monochromatic light of wavelength $500\, nm$ is used is

Two slits separated by $0.5 \,mm$ are illuminated by light of wavelength $500 \,nm$. The screen is at a distance of $120 \,cm$ from the slits. The phase difference between the interfering waves at a point $3 \,mm$ on the screen from the central bright fringe is ........... .

In Young's double slit experiment,the slits are horizontal. The intensity at a point $P$ on the screen is $\frac{I_0}{4}$,where $I_0$ is the maximum intensity. If the distance between the two slits $S_1$ and $S_2$ is $d = 2 \lambda$,then the value of $\theta$ is:

In a Young's double-slit experiment,interference fringes are obtained on a screen placed at a certain distance from the slits using monochromatic light. If the screen is moved towards the slits by $5 \times 10^{-2} \, m$,the fringe width changes by $3 \times 10^{-5} \, m$. If the distance between the slits is $10^{-3} \, m$,find the wavelength of the light.

The intensity at the maximum in a Young's double slit experiment is $I_0$. The distance between two slits is $d = 5\lambda$,where $\lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D = 10d$?

In an interference experiment,the third bright fringe is obtained using light of wavelength $700 \, nm$. What should be the wavelength of light to obtain the fifth bright fringe at the same point?