A transparent sheetof refractive index `1.5` is kept near one of the slits of the `YDSE` apparentus. The intenisity at the centre of the screen (where the central maximum was located before the introduction of the sheet) is half of the previous value. The minimum thickness of the sheet should be (wavelength of the monochromatic light used in the experiment is `6000 Å`) :

Options

A. `2 xx 10^(-7)m`
B. ` 7.5 xx 10^(-8) m`
C. (c ) `3 xx 10^(-7) m`
D. `2 xx 10^(-7) m`

Answer

C

Explanation

As per the question,
`I_(0) "cos"^(2)(phi)/(2) = (I_(0))/(2) :. "cos" (phi)/(2) = (1)/(sqrt(2)) = "cos" (pi)/(4)`
`(phi)/(2) = (pi)/(4) or phi = (pi)/(2)`.
As `phi = (2pi)/(lambda)(Delta x) = (pi)/(2)`
`:. Delta x = (lambda)/(4) = (mu - 1) t`
`t = (lambda)/(4(mu - 1)) = (6000 xx 10^(-10))/(4(1.5 - 1)) = 3 xx 10^(-7) m`