A double-slit apparatus is immersed in a liquid of refractive index 1.33. in It has slit separation of 1 mm and distance between the plane of slits and screen is 1.33 m. The slits are illuminated by a parallel beam or light whose wavelength in air is `6300 Å.`
a. Calculate the fringe width.
b. One of the slits of the apparauts is covered by a thin glass sheet of refractive index 1.53. Find the smallest thickness of the sheet to bring the adjacent minimum on the axix.

Explanation

KEY IDEA
If t is the thickness of the glass plate, the fringes are displaced by an amount given by
`Delta y=((mu_(g))/(mu_(1))-1)(tD)/(d)" "(35-24a)`
In order to bring the adjacent minimum on the axis (where we had the maximum before the sheet was introduced), the fringes must be displaced by half the fringe width, that is
`Delta y=(beta_(1))/(2)" "(35-24b)`
Calculation: Equating Eqs. 35-24a and 35-24b we have
`(beta_(1))/(2)=((mu_(g))/(mu_(1))-1)(tD)/(d)`
Substituting `beta_(1) = 6.3 xx 10^(4) m, mu_(g)= 1.53, mu_(1) = 1.33`, D = 1.33 m and `d= 1 xx 10^(-3)` m and solving we get
`t= 1.575 xx 10^(-6) m`.
Learn: By measuring this fringe shift, we can measure the thickness of very thin plates.