In a single slit diffraction pattern, a light of wavelength $6000 \AA$ is used. The distance between

Question

In a single slit diffraction pattern, a light of wavelength $6000 \AA$ is used. The distance between the first and third minima in the diffraction pattern is found to be 3 mm when the screen is placed 50 cm away from slits. The width of the slit is _______ $	imes 10^{-4} \mathrm{~m}$.\

Professor PC Thomas Classes · Answer

(2) : Wavelength, $\lambda=6000 \AA$; distance between first and third minima $=3 \mathrm{~mm}$\ $D=50 \mathrm{~cm}, d=$ ?\ Position of $n^{	ext {th }}$ maxima,\ $y=\frac{n D \lambda}{d}$\ So, $y_{1}=\frac{\lambda D}{d}, y_{3}=\frac{3 D \lambda}{d} ; \Delta y=y_{3}-y_{1}=\frac{2 \lambda D}{d}$\ $3 	imes 10^{-3}=\frac{2 	imes 6000 	imes 10^{-10} 	imes 0.5}{d}$ $$ d=\frac{2 	imes 6 	imes 0.5 	imes 10^{-7}}{3 	imes 10^{-3}}=2 	imes 10^{-4} \mathrm{~m} $$