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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 openings of a double slit apparatus. The interference pattern displaces itself through seven succesive maxima towards the side where the flake is placed. If wavelength of the diffracted light is `lambda=600nm`, then the thickness of

A

(a) `2100nm`

B

(b) `4200nm`

C

(c) `8400nm`

D

(d) None of these

Text Solution

Verified by Experts

The correct Answer is:
C
Shift`=beta/lambda(mu-1)t`
`implies7beta=beta/lambda(mu-1)timpliest=(7lambda)/((mu-1)`
`=(7xx600)/((1.5-1))=8400nm`.
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A2Z-WAVE OPTICS-Young'S Double Slit Experiment
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