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(C) The distance between consecutive bright or dark fringes is known as the fringe width,denoted by $\beta$.The formula for fringe width is $\beta = \

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$0.5$

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(C) The distance between consecutive bright or dark fringes is known as the fringe width,denoted by $\beta$.The formula for fringe width is $\beta = \frac{\lambda D}{d}$.Given:Wavelength $\lambda = 550 \, nm = 550 	imes 10^{-9} \, m$.Distance between slits $d = 1.10 \, mm = 1.10 	imes 10^{-3} \, m$.Distance to screen $D = 1 \, m$.Substituting the values:$\beta = \frac{550 	imes 10^{-9} 	imes 1}{1.10 	imes 10^{-3}}$$\beta = \frac{550}{1.10} 	imes 10^{-6} \, m$$\beta = 500 	imes 10^{-6} \, m = 0.5 	imes 10^{-3} \, m = 0.5 \, mm$.Thus,the correct option is $C$.

Young's double slit experiment is performed with light of wavelength $550 \, nm$. The separation between the slits is $1.10 \, mm$ and the screen is placed at a distance of $1 \, m$. What is the distance between consecutive bright or dark fringes in $mm$?

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