In Young's double-slit interference experimen | vedclass

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(B) Given: Distance between slits $d = 0.1 \ mm = 10^{-4} \ m$,distance of screen $D = 20 \ cm = 0.2 \ m$,and wavelength $\lambda = 5460 \ \mathring{A

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$0.16$

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(B) Given: Distance between slits $d = 0.1 \ mm = 10^{-4} \ m$,distance of screen $D = 20 \ cm = 0.2 \ m$,and wavelength $\lambda = 5460 \ \mathring{A} = 5460 	imes 10^{-10} \ m$.For the first dark fringe,the path difference is $\Delta x = \lambda / 2$.The angular position $	heta$ is given by the relation $d \sin 	heta = \Delta x = \lambda / 2$.Since $	heta$ is very small,$\sin 	heta \approx 	heta$ (in radians).$	heta = \frac{\lambda}{2d} = \frac{5460 	imes 10^{-10}}{2 	imes 10^{-4}} = 2730 	imes 10^{-6} \ rad$.To convert radians to degrees,multiply by $(180 / \pi)$:$	heta_{deg} = 2730 	imes 10^{-6} 	imes \frac{180}{3.14159} \approx 0.1564^\circ$.Rounding to two decimal places,we get $	heta \approx 0.16^\circ$.

In Young's double-slit interference experiment,the distance between two sources is $0.1 \ mm$. The distance of the screen from the sources is $20 \ cm$. The wavelength of light used is $5460 \ \mathring{A}$. What is the angular position of the first dark fringe in degrees?

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