In Fraunhofer diffraction,light of wavelength | vedclass

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(B) In Fraunhofer diffraction due to a single slit of width $a$,the angular position of the $n^{th}$ minima is given by $\sin 	heta = \frac{n \lambda

Vedclass · Accepted Answer

$0.18$

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(B) In Fraunhofer diffraction due to a single slit of width $a$,the angular position of the $n^{th}$ minima is given by $\sin 	heta = \frac{n \lambda}{a}$. For small angles,$	heta \approx \frac{n \lambda}{a}$.The angular width of the secondary maxima or minima is given by $\Delta 	heta = \frac{\lambda}{a}$.Given: $\lambda = 6328 	imes 10^{-10} \ m$ and $a = 0.2 	imes 10^{-3} \ m$.$\Delta 	heta = \frac{6328 	imes 10^{-10}}{0.2 	imes 10^{-3}} \ rad = 3.164 	imes 10^{-3} \ rad$.To convert radians to degrees,multiply by $\frac{180^{\circ}}{\pi}$:$\Delta 	heta = 3.164 	imes 10^{-3} 	imes \frac{180}{3.14159} \approx 0.18^{\circ}$.

In Fraunhofer diffraction,light of wavelength $6328 \ \mathring{A}$ is incident perpendicularly on a slit of width $0.2 \ mm$. The angular width of the secondary maxima or minima will be: (in $^{\circ}$)

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