Question

In a Young’s double slit experiment, two slits are located 1.5 mm apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 nm. If the 20 maxima of the double slit pattern are contained within the central maximum of the single slit diffraction pattern, then the width of each slit is \(x \times 10^{-3} \ cm\), where x-value is ___________.

Answer 1

Answer is : (15)

d = 1.5 mm

D = 2 m

\(\lambda = 400 \ nm\)

\(\frac{20 \lambda D}{d} = \frac{2\lambda}{a}\)

\(a = \frac{d}{10 D} = \frac{1.5}{10} \ mm\)

\(= \frac{150 \times 10^{-3 \ cm}}{10}\)

\(= 15 \times 10^{-3} \ cm\)