A double slit experiment is performed with light of wavelength 500 nm. A thin film of thickness 2 μ m and refractive index 1.5 is introduced in the path of the upper beam. The location of the central

Question

A double slit experiment is performed with light of wavelength 500 nm. A thin film of thickness 2 μ m and refractive index 1.5 is introduced in the path of the upper beam. The location of the central maximum

TestHub · Accepted Answer

The path difference introduced by the thin film is given by $Δ x = (μ - 1) t$

Substituting the given values:

$Δ x = (1.5 - 1) \times 2 \times 1 0^{- 6} = 1 0^{- 6}$ m

The number of fringes by which the central maximum shifts is $n = \frac{Δ x}{λ}$

$n = \frac{1 0 ^{- 6}}{500 \times 1 0 ^{- 9}} = 2$

Since the film is introduced in the path of the upper beam, the central maximum shifts towards the upper slit to compensate for the additional optical path.

Therefore, the central maximum shifts upward by two fringes.

Answer: shift upward by nearly two fringes

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