Width of the principal maximum on a screen at a distance of 50 cm from the slit having width 0.02 cm is `312.5 times 10^-3`cm. If waves were incident normally on the slit, then wavelength of the light from the source will be –

Width of the principal maximum on a screen at a distance of 50 cm from the slit having width 0.02 cm is 312.5xx10^(-3)cm . If waves were incident normally on the slit, then wavelength of the light from the source will be

Width of the principal maximum on a screen at a distance of 50 cm from the slit having width 0.02 cm is 312.5xx10^(-3)cm . If waves were incident normally on the slit, then wavelength of the light from the source will be

Light of wavelength lambda is incident on a slit. First minima of the diffraction pattern is found to lie at a distance of 6 mm from the central maximum on a screen placed at a distance of 2 m from the slit. If slit width is 0.2 mm, then wavelength of the light used will be :

Light of wavelength 6328 Å is incident normally on a slit of width 0.2 mm. Angular width of the central maximum on the screen will be :

Light of wavelength 6328 Å is incident normally on a slit of width 0.2 mm. Angular width of the central maximum on the screen will be :

Light of wavelength 6 xx10^(-5) cm falls on screen at a distance of 100 cm from a narrow slit. Find the width of the slit if the first minima lies 1mm on either side of the central maximum

Light of wavelength 6 xx10^(-5) cm falls on screen at a distance of 100 cm from a narrow slit. Find the width of the slit if the first minima lies 1mm on either side of the central maximum

In a double experiment, the distance between the slit is 1 mm and screen is 25 cm away from the slits. The wavelength of light is 6000Å . The width of the fringe on the screen is