In a Young's double slit experiment, two slits are located 1.5 mm apart. The distance of screen from

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 $x 	imes 10^{-3} \mathrm{~cm}$, where $x$-value is $\_\_\_\_$ .\

Professor PC Thomas Classes · Answer

(15) : $d=1.5 \mathrm{~mm}, D=2 \mathrm{~m}, \lambda=400 \mathrm{~nm}$ Width of 20 maxima of double slit = Width of central maxima of single slit. $$ \begin{aligned} & \frac{20 \lambda D}{d}=\frac{2 D \lambda}{a} ; \frac{10}{d}=\frac{1}{a} \ & a=\frac{d}{10}=\frac{1.5 	imes 10^{-3}}{10}=15 	imes 10^{-3} \mathrm{~cm} \ 	herefore \quad & x=15 \end{aligned} $$