A block o plastic having a thin air cavity (whose thickenss is comparable to wavelength of light waves) is shown in fig. The thickness of air cavity (which can be considered as air wedge for interference pattern) is varying linearly from one end to other as shown.
A broad beam of monochromatic light is incident normally from the top of the plastic box. Some ligth lis reflected back above and below the cavity .The plastic layers than wavelength of incident light . An observer when looking down from top sees an interference pattern consisting of eight dark fringes and seven bright fringe along the wedge. Take wavelength of incident light in air as `lambda_(0)` and refractive index of plastic as `mu`
Assume that the thickness of the ends of air cavity are such that formation of fringe takes place there.

Determine of difference `L_(1) - L_(2) = Delta L)` in terms of `lambda_(0)`.

Options

A. `(4 lambda_(0))/(mu)`
B. `(7 lambda_(0))/(2 mu)`
C. `(3 lambda_(0))/(mu)`
D. None of the above

Answer

b

Explanation

Form the theory mentioned in passage, at the ends of cavity fringes will from, and as number of dark fringes is greater than the number of bright fringes so the ends will be location of dark fringes. Thickness of the cavity at a distance x from the left end would be
`t = L_(2) + (L_(1) - L_(2))/(L) x`
For left end,
`2 mu L_(2) = (n - 7) lambda_(0)`
`L_(1) - L_(2) = (7 lambda_(0))/(2 mu)`.