Light of wavelength $6000 \AA$ falls on a single slit of width 0.1 mm . The second minimum will be f

Question

Light of wavelength $6000 \AA$ falls on a single slit of width 0.1 mm . The second minimum will be formed for the angle of diffraction of\

Professor PC Thomas Classes · Answer

d) Here, wavelength $=6000 \mathrm{~A}^{0}=6000 	imes 10^{-10} \mathrm{~m}$ slit width $1=0.1 \mathrm{~mm}=0.1 	imes 10^{-3} \mathrm{~m}$ In case of diffraction at single slit, the position of minima is given by\ $\operatorname{asin} 	heta_{\mathrm{n}}=\mathrm{n} \lambda$\ where $\mathrm{n}=1,2,3, \ldots$\ For second minimum, $\mathrm{n}=2$\ $	herefore a \sin 	heta_{2}=2 \lambda$\ For small angle $\sin 	heta_{2} 	heta_{2} 	herefore 	heta_{2}=\frac{2 \lambda}{a}$\ Substituting the values, we get\ $	herefore 	heta_{2}=\frac{2 	imes 6000 	imes 10^{-10}}{0.1 	imes 10^{-3}}$\ $=0.012$ radian\