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A flake of glass (refractive index 1.5) ...

A flake of glass (refractive index 1.5) is placed over one of the opening of a double-slit apparatus. The interference pattern displaced itself through seven successive maxima toward the side where the flake is placed. If wavelength of the light is `lambda = 600 nm`, then the thickness of the flake is

A

2100 nm

B

4200 nm

C

8400 nm

D

none of above

Text Solution

Verified by Experts

The correct Answer is:
c

(Shift) `= 7` (Position of central maxima)
`implies (mu - 1)t (D)/(d) = 7 ((lambda D)/(d))`
`:.t = (7 lambda)/(mu - 1)`
` implies t = (7(600))/(0.5)`
`implies t = 8400 nm`
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