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.
Change is fringe width produced due to chanbe in wavelength is

Options

A. `6.22xx10^(-5)m`
B. `-6.22xx10^(-5)m`
C. `-5.71xx10^(-5)m`
D. `5.71xx10^(-5)m`

Answer

C

Explanation

In part (i), shifting of 5 bright fringes was equal to `10^(-3)m`. Which implies that
`5W_(red) = 10^(-3)m` [Here W = Fringe width]
`W_(red) = (10^(-3))/(5)m = 0.2xx10^(-3)m`
Now since w = `(lambda D)/(d) or w mu 1`
`(omega_("green"))/(omega_("red")) = (lambda_("green"))/(lambda_("red"))`
`W_("green") = W_("red") (lambda_("green"))/(lambda_("red")) = (0.2xx10^(-3)) ((5xx10^(-7))/(7xx10^(-7)))`
`W_("green") = 0.143xx10^(-3)m`
`Delta w = w_("green") - w_("red") = (0.143-0.2)xx10^(-3)m`
`Delta w = -5.71xx10^(-5)m`