A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observe on screen 1 m away. It is observed that the first minimum is at a distance of `2.5 mm` from the centre of the screen. Find the width of the slit.

Explanation

Here, wavelength, `lambda = 500 nm = 5 xx 10^(-7) m`,
Distance between the slit and the screen , `D = 1 m`
Distance of first minimum from the central
maximum, `y = 2.5 mm = 2.5 xx 10^(-3) m`
Size of the aperture , a = ?
For the first secondary minimum,
`sin theta = (lambda)/a = y/D`
`:. a = (lambda D)/lambda = (5 xx 10^(-7) xx 1)/(2.5 xx 10^(-3)) = 2 xx 10^(-4) m`
`=0.2 mm`.