A mica strip and a polysterence strip are fitted on the two slite of a double slit apparatus. The thickness of the strips is 0.50 mm and the separation between the slits is 0.12 cm. The refractive index of mica and polysterene are 1.58 and 1.55 respectively for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen a distance one meter away. (a) What would be the fringe- width ? (b) At what distance from the centre will the first maximum be located ?

Answer

A, B, C, D

Explanation

Given that, `t_1 = t_2 = t =0.5`
`mu_m = 0.5 xx 10^(-3)m`
`mu_m = 1.58 and mu_p = 1.55`
`lambda = 590 nm`
`= 590 xx 10^(-9) m,`
`d = 0.12 cm = 12 xx 10^(-4)m`
D =1m
(a) Fringe width ` =(D lambda)/(d)`
` =(1xx 590xx 10^(-9))/(12 xx 10^(-9))`
`=4.91 xx 10^(-4)m`
(b) When both the strip are fitted, the optical path changes by
`Delta x = (mu_0 -1 ) t - (mu_p - 1)t`
`=(mu_m - mu_p)t`
`=(1.58 - 1.55) xx (0.5) (10^(-3))`
`=(0.15) xx 10^(-3)m`
So, no of fringes shifted
` = (0.015 xx 10^(-3))/(590 xx 10^(-9)) = 25.43`
`rArr` There are 25 fringes and `0.43th` of a fringe.
`rArr` There are 13 bright fringes and 12 dark fringes and `0.43th` of a dark fringe. So, position of first maximum on both sides will be given by,
`:. x = (0.43) xx 4.91 xx 10^(-4)`
= 0.021 cm
`x ' = (1 - 0.43) xx 4.91 xx 10^(-4)`
= 0.028 cm
[Since, fringe width `= 4.91 xx 10^(-4)m]`