A double slit experiment is performed with light of wavelength 500 nm- A thin film of thickness 2 - m and refractive index 1-5 is introduced in the path of the upper beam- The location of the central maximum will

The correct option is B Shift downward by nearly two fringesP = S2P #160; #160;[S1P+ #160; #181;t #160; #160;t] = S2P #160; #160;S1P #160; ...

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Standard XII

Physics

Huygens' Principle

A double slit...

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 will

Remain unshifted

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Shift downward by nearly two fringes

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Shift upward by nearly two fringes

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Shift downward by ten fringes

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Solution

The correct option is B Shift downward by nearly two fringes

P = S2P - [S1P+ µt - t] = S2P - S1P - (µ - 1)t = y.d/D - (µ - 1)

⇒  nth fringe is shifted by  Δy = D(µ-1)t/d = w/λ (µ-1)t

As each fringe width = w,

The number of fringes that will shift = total fringe shift/fring width

(w/λ(µ-1)t)/w = (µ-1)t/λ = (1.5-1) x .2 x 10-5m / 500 x 10-9 = 2

Hence, will shift downward by nearly 2 fringes

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

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 will