In a YDSE fringes are observed by using light of wavelength 4800 Å, if a glass plate ( $μ$ = 1.5) is introduced in the path of one of the waves and another plate( $μ$ =1.8) of same thickness is introduced in the path of the other wave. The central fringe takes the position of fifth bright fringe. The thickness of each plate will be(in micron)

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0.8

800

answer is A.

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Detailed Solution

$Shift due to one plate {Δx}_{1} = \frac{β}{λ} (μ_{1} - 1)$

$Shift due to another path {Δx}_{2} = \frac{β}{λ} (μ_{2} - 1) t$

$Net shift Δ x = Δ x_{2} - Δ x_{1} = \frac{β}{λ} (μ_{2} - μ_{1}) t$

$Also it is given that Δ x = 5 β$

$Hence, 5 β = \frac{β}{λ} (μ_{2} - μ_{1}) t$

$\Rightarrow t = \frac{5 λ}{(μ_{2} - μ_{1})} = \frac{5 \times 4800 \times 10^{- 10}}{(1.8 - 1.5)} = 8 \times 10^{- 6} m = 8 μ m$

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