1.

ABC is a spherical wavefront centred at 'O' symmetric about BE is incident on slits S_(1) and S_(2) .BS_(1) = 3lambda , S_(1)S_(2) = 4 lambda , BO = 6lambda ,S_(1)E = 128 lambdaand lambdais the wavelength of incident light wave. A mica sheet of refractive index 1.5 is pated on S_(2). Find the minimum value of thickness of mica sheet for whhichcentral fringe forms at E ?

Answer»

`(31lambda)/(8)`
`(15lambda)/(8)`
`(5lambda)/(8)`
`(7lambda)/(8)`

SOLUTION :LET `S_(1)^(1)`and `S_(2)^(1)` are the points on the wavefront where PERPENDICULARS can be DRAWN from `S_(1)` and `S_(2)` .Four central FRINGE formed at E the path difference should be zero.
`S_(2)S_(2)+(mu - 1)t+S_(2)E = S_(1)^(1)S_(1)+S_(1)E`
`implies(OS_(2)^(1)-OS_(2))+(mu-1)t+sqrt(D^(2)+d^(2)) = (OS_(1)^(1)-OS_(1))+D`
`implies(mu-1)t = (OS_(2)-OS_(1))+(sqrt(D^(2)+d^(2))-D)`
`implies(mu-1)t = 2lambda - [D(1+(d^(2))/(D^(2)))^(1//2)-D]`
`implies2lambda - [D(1+(d^(2))/(2D^(2)))-D] implies 2 lambda - [D+(d^(2))/(2D)-D]`
`(mu-1)t=2lambda - (d^(2))/(2D) ((3)/(2) - 1)t = 2xxlambda - (16lambda^(2))/(2xx128lambda)`
`(1)/(2)t = 2lambda - (lambda)/(16) (1)/(2)t = (31lambda)/(16) = (31lambda)/(8)`
Thickness of mica sheet `t = (31lambda)/(8)`


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