In a Young's double slit experiment the intensity at a point where the path difference is $${\lambda \over 6}$$ ( $$\lambda $$ being the wavelength of light used ) is $$I$$. If $${I_0}$$ denotes the maximum intensity, $${I \over {{I_0}}}$$ is equal to

Two point white dots are $$1$$ $$mm$$ apart on a black paper. They are viewed by eye of pupil diameter $$3$$ $$mm.$$ Approximately, what is the maximum distance at which these dots can be resolved by the eye? [ Take wavelength of light $$=500$$ $$nm$$ ]

A Young's double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is

If $${I_0}$$ is the intensity of the principal maximum in the single slit diffraction pattern, then what will be its intensity when the slit width is doubled?