A double slit apparatus is immersed in a liquid of refractive index 1.33. It has slit and the screen 1 mm. The slits are illuminated by a parallel beam of light whose wavelength in air is `6300 Å`
a. calculate the fringe width.
b. One of the slits of the apparatus is covered by a thin glass sheet of refractive index 1.53. Find the smallest thickness of the sheet to bring athe adjacent minima on the axis.

Answer

A, C, D

Explanation

(i) `_m^amu=(lamda_a)/(lamda_m)implieslamda_m=(lamda_a)/(`_m^2mu)`
`:'` Fringe width `=(lamda_aD)/(`_m^amud)=(6300xx10^(-10)xx1.33)/(1.33xx10^(-3))`
`=6.3xx10^(-4)m`
(ii) KEY CONCEPT: The shift of fringes when one slit is covered with thin glass sheet is
`=(Dt)/(d)[(g^mu)/(g^mu)-1]`
where, t=thickness of glass sheet. The shift has to be such that the minima shifts to the sxis. For this the shifting of the fringes should be `(beta)/(2)` where `beta` is fringe width.
`:'` `(Dt)/(d)[(g^mu-m^mu)/(m^mu)]=(beta)/(2)`
`implies``t=(betadm^mu)/(2(g^mu-m^mu)xxD)=(6.3xx10^(-4)xx10^(-3)xx1.33)/(2(1.53-1.33)xx1.33)`
`=15.75xx10^(-7)m=1.575xx10^(-6)m`