In Young's experiment, the source in red light of wavelength `7 xx 10^(-7)` m. When a thin glass plate of refractive index. 1.5 at this wavelength is put in the path of one of the interfering beams, the central bright fringe shifts by `10^(3)` m to position previously occpied by the 5th bright fringe. Find the thickness of the plate.
When the source is now changed to green light of wavelength `5 xx 10^(-7)` m, the central fringe shifts to position initially occupied by the 6th bright fringe due to red light, Find the refractive index of the glass for the green light. Also, estimate the change in fringe width due to the change in wavelength.

Options

A. `2.6 mu m`
B. `1.6 mu m`
C. `1.2 mu m`
D. `2.2 mu m`

Answer

b

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.