Home
Class 12
PHYSICS
In YDSE, the slits separation is 0.6mm a...

In YDSE, the slits separation is `0.6mm` and the separation between slit and screen is `1.2m`. The wavelength of light used is `4800 overset(@)AA` . The angular fringe width is `Pxx10^(-4)` radian. Find the value of P ?

Promotional Banner

Similar Questions

Explore conceptually related problems

In Young's double slit experiment, slit separation is 0.6 mm and the separation between slit and screen is 1.2m . The angular width is (the wavelength of light used is 4800Å )

In YDSE experiment separation between plane of slits and screen is 1 m . Separation between slits is 2 mm . The wavelength of light is 500 nm . The fringe width is

In YDSE, separation between slits is 0.15 mm, distance between slits and screen is 1.5 m and wavelength of light is 589 nm, then fringe width is

In YDSE, separation between slits is 0.15 mm, distance between slits and screen is 1.5 m and wavelength of light is 589 nm, then fringe width is

In YDSE, separation between slits is 0.15 mm, distance between slits and screen is 1.5 m and wavelength of light is 589 nm, then fringe width is

In YDSE wavelength of light used is 500 nm and slit width is 0.05 mm . then the angular fringe width will be.

In YDSE wavelength of light used is 500 nm and slit width is 0.05 mm . then the angular fringe width will be.

In YDSE apparatus screen is placed at distance of 1m, seperation between slit is 1 mm and fringe width is 6mm. Find wavelength of light in nm.

In a Young’s double slit experiment, the slit separation is 0.2 cm , the distance between the screen and slit is 1 m . Wavelength of the light used is 5000 Å . The distance between two consecutive dark fringes (in mm ) is