In YDSE, the sources is red ligth of wavelength `7 xx 10^(-7) m`. When a thin glass plate of refractive index 1.5 is put in the path of one of the interfering beams, the central bright fringe shifts by `10^(-3)` m to the position previously occupied by the 5th bright fringe.
What is the thickness of the plate?

Options

A. `- 0.57 xx 10^(-4) m`
B. `- 0.47 xx 10^(-4) m`
C. `- 0.37 xx 10^(-4) m`
D. `- 0.27 xx 10^(-4) m`

Answer

a

Explanation

Introduction of a glass plate shifts the central maxiama by
`Delta x = (D(mu - 1) t)/(d) = (beta(mu - 1)t)/(lambda)`
As the central maxima shifts by five fringes, therefore
`(beta_(R) (mu_(R) - 1) t)/(lambda_(R)) = 5 beta_(R)`
`t = (5 lambda_(R))/((mu_(R) - 1)t) = (5 xx 7 xx 10^(-6))/((1.5 - 1)) = 7 mu m`
Similarly, when green light is used. the central maxima shifts by six fringes, therefore
`(beta_(G) (mu_(G) - 1)t)/(lambda_(G)) = 6 beta_(G)`
Dividing Eq. (i) by Eq.(ii), we get
`((mu_(R) - 1))/((mu_(G) - )) = (5)/(6) implies mu_(G) - 1 = (6)/(5) (1.5 -1)`
`mu_(G) = 1.6`
As given in the problem,
`5 beta_(R) = 110^(-3) implies beta_(R) = 2 xx 10^(-4)`
`:. (beta_(G))/(beta_(R)) = ((lambda_(G) D // d))/((lambda_(R) D // d)) = (lambda_(G))/(lambda_(R)) = (5)/(7)`
`implies (beta_(G))/(beta_(R)) - 1 = (5)/(7) - 1 = - (2)/(7)`
`:. Delta beta = beta_(G) - beta_(R) = - (2)/(7) xx beta_(R) = - 0.57 xx 10^(-4) m`
The minus sign denotes that fringe width will decrease when green light is used.