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In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength λ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is

(a)  2λ

(b)  2λ/3

(c)  λ/3 

(d)  λ 

1 Answer

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Best answer

Correct option  (a) 2λ

Explanation:

 Path difference due to slab (μ −1) t . For maximum intensity path difference should be equal nλ . Thus (μ − 1) t  = nλ

For minimum thickness t of plate, n should be minimum i.e. n = 1

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