The waves of 600 mu m wavelength are inciden | vedclass

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(B) For single slit diffraction,the condition for the $n^{th}$ minima is given by $d \sin 	heta = n \lambda$.For the first minima,$n = 1$,so the cond

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$\frac{\pi}{6}$

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(B) For single slit diffraction,the condition for the $n^{th}$ minima is given by $d \sin 	heta = n \lambda$.For the first minima,$n = 1$,so the condition becomes $d \sin 	heta = \lambda$.Given wavelength $\lambda = 600 \ \mu m = 600 	imes 10^{-6} \ m = 6 	imes 10^{-4} \ m$.Given slit width $d = 1.2 \ mm = 1.2 	imes 10^{-3} \ m = 12 	imes 10^{-4} \ m$.Substituting the values into the formula: $\sin 	heta = \frac{\lambda}{d} = \frac{6 	imes 10^{-4}}{12 	imes 10^{-4}} = \frac{1}{2}$.Since $\sin 	heta = \frac{1}{2}$,the angle $	heta = \arcsin(0.5) = \frac{\pi}{6} 	ext{ radians}$.

The waves of $600 \ \mu m$ wavelength are incident normally on a slit of $1.2 \ mm$ width. The value of diffraction angle corresponding to the first minima will be (in radian) -

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