In Young's double slit experiment, using monochromatic light, the fringe patternn shifts by a certain distance on the screen, when a transparent sheet of thickness 't' and refractive index `mu` is introduced in the path of one of the interfering waves. the sheet is then removed and the distance between the screen and the slits is doubled. it is found that the distance between the successive maxima (or minima) now is the samme as the observed fringe shift when the sheet was introduced. what is the wavelength of light used?

Options

A. `lamda=(mu-1)t`
B. `lamda=((mu-1)t)/(2)`
C. `lamda=(mu+1)t`
D. `lamda=((mu+1)t)/(2)`

Answer

B

Explanation

The fringe system is shifted through a distance a, given by
`s=(mu-1)t*(X)/(lamda)=(mu-1)t*(lamdaD)/(d)xx(1)/(lamda)`
`=(mu-1)t*(D)/(d)` . . . (1)
Where the fringe width `X=(lamdaD)/(d)`
when the plate is removed and D is doubled, the new fringe width becomes
`X'=(2lamdaD)/(d)` . . . (2)
It is given that `s=X'`
`therefore (mu-1)t(D)/(d)=(2lamdaD)/(d)" "therefore lamda=((mu-1)t)/(2)`