Light of wavelength 550, nm falls normally on | vedclass

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(D) For single slit diffraction,the condition for the $n^{th}$ minima is given by $a \sin 	heta = n\lambda$,where $a$ is the slit width,$\lambda$ is

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$\frac{\pi}{6}$

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(D) For single slit diffraction,the condition for the $n^{th}$ minima is given by $a \sin 	heta = n\lambda$,where $a$ is the slit width,$\lambda$ is the wavelength,and $	heta$ is the angular position.Given: $\lambda = 550\, nm = 550 	imes 10^{-9}\, m$,$a = 22.0 	imes 10^{-5}\, cm = 22.0 	imes 10^{-7}\, m$,and $n = 2$ for the second minima.Substituting the values into the formula:$a \sin 	heta = 2\lambda$$\sin 	heta = \frac{2\lambda}{a} = \frac{2 	imes 550 	imes 10^{-9}}{22.0 	imes 10^{-7}}$$\sin 	heta = \frac{1100 	imes 10^{-9}}{22.0 	imes 10^{-7}} = \frac{1100}{2200} = 0.5$Since $\sin 	heta = 0.5$,we have $	heta = \arcsin(0.5) = \frac{\pi}{6}\, rad$.

Light of wavelength $550\, nm$ falls normally on a slit of width $22.0 \times 10^{-5}\, cm$. The angular position of the second minima from the central maximum will be (in radians)

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