In a Young's double slit experiment with slit | vedclass

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(A) For bright fringes (maxima) in a Young's double slit experiment,the condition is $d \sin 	heta = n \lambda$,where $n$ is the order of the fringe.

Vedclass · Accepted Answer

$625\, nm, 500\, nm$

Vedclass · Answer

(A) For bright fringes (maxima) in a Young's double slit experiment,the condition is $d \sin 	heta = n \lambda$,where $n$ is the order of the fringe.Since $	heta$ is small,$\sin 	heta \approx 	heta = \frac{1}{40}\, rad$.Given $d = 0.1\, mm = 10^{-4}\, m$.The condition becomes $d 	heta = n \lambda$,so $\lambda = \frac{d 	heta}{n}$.Substituting the values: $\lambda = \frac{10^{-4} 	imes (1/40)}{n} = \frac{10^{-4}}{40n} = \frac{10^{-5}}{4n} = \frac{2500\, nm}{n}$.For $\lambda$ to be in the visible range ($380\, nm$ to $740\, nm$):If $n=4$,$\lambda = \frac{2500}{4} = 625\, nm$.If $n=5$,$\lambda = \frac{2500}{5} = 500\, nm$.Both $625\, nm$ and $500\, nm$ lie within the visible range. Thus,the wavelengths are $625\, nm$ and $500\, nm$.

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