In Young's double slit experiment, two slits S 1 and S 2 are 'd' distance apart and the separation from slits to screen is D . Now if two transparent slabs of equal thickness 0.1 mm

Question

In Young's double slit experiment, two slits S 1 ​ and S 2 ​ are 'd' distance apart and the separation from slits to screen is D . Now if two transparent slabs of equal thickness 0.1 mm but refractive index 1.51 and 1.55 are introduced in the path of beam ( λ = 4000 Å) from S 1 ​ and S 2 ​ respectively. The central bright fringe spot will shift by ___ number of fringes.

TestHub · Accepted Answer

The net path difference introduced at the center of the screen due to the two transparent slabs is given by:

$Δ x = (μ_{2} - 1) t - (μ_{1} - 1) t = (μ_{2} - μ_{1}) t$

The number of fringes by which the central bright fringe shifts is:

$n = \frac{Δ x}{λ} = \frac{( μ _{2} - μ _{1} ) t}{λ}$

Substituting the given values, $μ_{1} = 1.51$ , $μ_{2} = 1.55$ , $t = 0.1 \times 1 0^{- 3}$ m, and $λ = 4000 \times 1 0^{- 10}$ m:

$n = \frac{( 1.55 - 1.51 ) \times 1 0 ^{- 4}}{4 \times 1 0 ^{- 7}}$

$n = \frac{0.04 \times 1 0 ^{- 4}}{4 \times 1 0 ^{- 7}}$

$n = \frac{4 \times 1 0 ^{- 6}}{4 \times 1 0 ^{- 7}} = 10$

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