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

The distance of the nth minimum from the centre of the screen is, `x_(n) = (n D lambda)/(a)`
Where, D = distance of slit from screen : `lambda` = wavelength of the light : a = width of the slit
For first minimum, `" "n = 1`
Thus, `2.5 xx 10^(-3) = (1(1)(500 xx 10^(-9)))/(a)`
`a = 2 xx 10^(-4) m`.