In Young's double slit experiment, the slit width is 1 mm and screen is at a distance 2 m from slits. Light of 600 nm is used. If a thin glass plate of refractive index 1.5 of thickness 0.06 mm i

Question

In Young's double slit experiment, the slit width is 1 mm and screen is at a distance 2 m from slits. Light of 600 nm is used. If a thin glass plate of refractive index 1.5 of thickness 0.06 mm is placed over one of the slits then there will be a lateral displacement of the fringes by

TestHub · Accepted Answer

The lateral displacement of the fringes due to the introduction of a thin glass plate is given by the formula:

$Δ y = \frac{D}{d} (μ - 1) t$

Given values are:
$D = 2$ m
$d = 1$ mm $= 1 0^{- 3}$ m
$μ = 1.5$
$t = 0.06$ mm $= 6 \times 1 0^{- 5}$ m

Substituting these values into the formula:

$Δ y = \frac{2}{1 0 ^{- 3}} (1.5 - 1) \times 6 \times 1 0^{- 5}$

$Δ y = 2000 \times 0.5 \times 6 \times 1 0^{- 5}$

$Δ y = 1000 \times 6 \times 1 0^{- 5}$

$Δ y = 6 \times 1 0^{- 2}$ m

$Δ y = 6$ cm

Answer: $6$ cm

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