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In a Young's double-slit experiment the separation?
In a Young's double-slit experiment the separation distance y between the second-order bright fringe and the central bright fringe on a flat screen is 0.0191 m, when the light has a wavelength of 425 nm. Assume that the angles are small enough so that sin theta is approximately equal to tan theta. Find the separation y when the light has a wavelength of 601 nm.
1 Answer
- ukmudgalLv 61 decade agoFavorite Answer
If slit separation is ' d ',screen distance is ' D 'then
path difference is d sin theta = n *wavelength( lambda)
Also tan theta =y / D
as angle theta is small,sin theta= tan theta
therefore, y=n * lambda * D/d
As n , d and D remain same, y is directly proportional to (wavelength) lambda,
y(2) / y (1) =lambda (2) / lambda ( 1 )
y ( 2 )= y ( 1 ) lambda (2) / lambda ( 1 )
=0.0191 *601 / 425 = 0.027 m
=0.027 m