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(A) The angular position of the $n^{th}$ dark fringe is given by $	heta_n = (2n-1) \frac{\lambda}{2d}$.For the first dark fringe $(n=1)$,the angular

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$0.45$

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(A) The angular position of the $n^{th}$ dark fringe is given by $	heta_n = (2n-1) \frac{\lambda}{2d}$.For the first dark fringe $(n=1)$,the angular position is $	heta = \frac{\lambda}{2d}$.Given: $\lambda = 5000 \, \mathring{A} = 5 	imes 10^{-7} \, m$,$d = \frac{0.1}{\pi} \, mm = \frac{0.1}{\pi} 	imes 10^{-3} \, m = \frac{10^{-4}}{\pi} \, m$.Substituting the values:$	heta = \frac{5 	imes 10^{-7}}{2 	imes (10^{-4} / \pi)} = \frac{5 	imes 10^{-7} 	imes \pi}{2 	imes 10^{-4}} = 2.5 	imes 10^{-3} 	imes \pi 	ext{ radians}$.To convert radians to degrees,multiply by $\frac{180}{\pi}$:$	heta_{deg} = (2.5 	imes 10^{-3} 	imes \pi) 	imes \frac{180}{\pi} = 2.5 	imes 10^{-3} 	imes 180 = 0.45^o$.

In a Young's double-slit experiment,the distance between the two sources is $0.1 / \pi \, mm$. The distance between the source and the screen is $25 \, cm$. The wavelength of light is $5000 \, \mathring{A}$. The angular position of the first dark fringe is ........$^o$.

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