In YDSE distance between two slits is d=4λ and separation between screen and slits is D (D>>d). Distance between central maxima and second maxima on the screen is (λ is wavelength of light used :.

In YDSE ,how many maxima can be obtained on the screen if wavelength of light used is 200nm and d=700nm :

In Young's experiment the distance between two slits is d/3 and the distance between the screen and the slits is 3D. The number of fringes in 1/3 metre on the screen, formed by monochromatic light of wavelength 3lambda , will be:

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, water is filled in the space between the slits and screen. Then

A beam of light consisting of two wavelengths 6500Å and 5200Å is used to obtain interference fringes in YDSE. The distance between slits is 2mm and the distance of the screen form slits is 120 cm. What is the least distance from central maximum where the bright due to both wavelengths coincide?

In young's double slit experiment, distance between the slit S_1 and S_2 is d and the distance between slit and screen is D . Then longest wavelength that will be missing on the screen in front of S_1 is

In a YDSE, distance between the slits and the screen is 1m, separation between the slits is 1mm and the wavelength of the light used is 5000nm . The distance of 100^(th) maxima from the central maxima is:

In Young's double-slit experiment, the separation between the slits is d, distance between the slit and screen is D (D gt gt d) , In the interference pattern, there is a maxima exactly in front of each slit. Then the possilbe wavelength(s) used in the experiment are

In a YDSE using monochromatic visible light, the distance between the plate of slits and the screen is 1.7 m. At point P on the screen which is directly in front of the upper slit, maximum path is observed. Now, the screen is moved 50 cm closer to the plane of slits. Point P now lies between third and fouth minima above the central maxima and the intensity at P in one-fourth of the maxima intensity on the screen. ltbr. Find the wavelength of light if the separation of slits is 2 mm.