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Question

(C) The fringe width (distance between two consecutive fringes) is given by the formula: $\beta = \frac{\lambda D}{d}$.Given values are:$\lambda = 500

Vedclass · Accepted Answer

$0.33$

Vedclass · Answer

(C) The fringe width (distance between two consecutive fringes) is given by the formula: $\beta = \frac{\lambda D}{d}$.Given values are:$\lambda = 500 \ nm = 500 	imes 10^{-9} \ m = 5 	imes 10^{-7} \ m$$D = 2 \ m$$d = 3 \ mm = 3 	imes 10^{-3} \ m$Substituting these values into the formula:$\beta = \frac{5 	imes 10^{-7} 	imes 2}{3 	imes 10^{-3}}$$\beta = \frac{10 	imes 10^{-7}}{3 	imes 10^{-3}}$$\beta = \frac{10}{3} 	imes 10^{-4} \ m$$\beta = 3.33 	imes 10^{-4} \ m$$\beta = 0.333 	imes 10^{-3} \ m = 0.33 \ mm$.Therefore, the correct option is $C$.

The distance between two slits is $3 \ mm$ and the screen is placed at $2 \ m$ distance. When blue-green light of wavelength $500 \ nm$ is used, then the distance between two consecutive fringes will be (in $mm$)?

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