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In Young's double slit experiment the two slits are covered by slabs of same thickness but refractive index 1.4 and 1.7. If the slit to screen separation is 1 m and slits are at 1 mm separation using a coherent source of wavelength 4000 $\overset{o}{A}$ and the central fringe shifts to the 3rd bright fringe position, then

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slabs are thickness 4 $μ m$

shift will be towards slab of index 1.4 by 1.2 mm

shift will be towards slab of index 1.7 by 1.2 mm

slabs are thickness 4 $\overset{o}{A}$

answer is A, C.

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Detailed Solution

Shift due to the first place is $x_{1} = \frac{β}{λ} (μ_{1} - 1) t$

Fringe width $β = \frac{λ D}{d} = \frac{4000 \times 10^{- 7} \times 10^{3}}{1} m m = 0.4 m m$

Next shift is towards the plate with higher $μ$ , so

$∆ x = 3 β = \frac{β}{λ} (μ_{2} - μ_{1}) t \Rightarrow t = \frac{3 λ}{μ_{2} - μ_{1}} = \frac{3 (4000 \times 10^{- 10})}{1.7 - 1.4} = 4 μ m S i f t o f c e n t r a l m a x i m a = 3 β = 1.2 m m$

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