A parallel beam of light of wavelength `4000 Å` passes through a slit of width `5 xx 10^(-3) m`. The angular spread of the central maxima in the diffraction pattern is

Light of wavelength 600 nm is incident normally on a slit of width 0.2 mm. The angular width of central maxima in the diffraction pattern is

Light of wavelength 600 nm is incident normally on a slit of width 0.2 mm. The angular width of central maxima in the diffraction pattern is

Find the angular width of the central maxima of Fraunhofer diffraction pattern due to single slit.

A parallel beam of light of wavelength 6000 Å gets diffracted by a single slit of width 0.3 mm.The angular position of the first minima of diffracted light is

A narrow slit is illuminated by a parallel beam of monochromatic light of wavelength lambda_(0) equals to 6000 A^(0) and the angular width of the central maxima in the resulting diffraction pattern is measured.When the slit is next illuminated by light of wavelength lambda ,the angular width decreases to half.Calculate the value of the wavelength lambda .

A parallel beam of light of wavelength 6000Å gets diffracted by a single slit of width 0.3 mm. The angular position of the first minima of diffracted light is :

A parallel beam of light of wavelength 6000Å gets diffracted by a single slit of width 0.3 mm. The angular position of the first minima of diffracted light is :