Light of wavelength 6000 A ~ … is incident on a thin glass plate of μ = 1.5 such that the angle of refraction into the plate is 6 0 ∘ . Calculate the smallest thickness of the plate which will make da

Question

Light of wavelength 6000 A ~ … is incident on a thin glass plate of μ = 1.5 such that the angle of refraction into the plate is 6 0 ∘ . Calculate the smallest thickness of the plate which will make dark fringe by reflected beam interference.

TestHub · Accepted Answer

For reflected light, condition of destructive interference, 2 μ t cos r − 2 λ ​ = ( 2 n + 1 ) 2 λ ​ ⇒ 2 μ t cos r = ( 2 n + 1 ) 2 λ ​ + 2 λ ​ For minimum thickness of the plate, n = 0 , So, 2 μ t cos r = λ ​ ⇒ 2 × 1.5 × t × cos 6 0 ∘ = 6000 A ~ … ⇒ t = 1.5 6000 ​ = 4000 A ~ … ​

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