In a Young's double-slit experiment the separation?

Question

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.

Answer 1

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