In a certain double slit experimental arrangement, interference fringes of width 1 mm each are observed when light of wavelength 5000 A^o is used. Keeping the setup unaltered, if the source is replaced by another of wavelength 6000 A^o, the fringe width will be

1.2 mm

0.5 mm

1 mm

1.5 mm

Solution

1.2 mm

Fringe width

β = $\frac{λ D}{d}$

Since, D and d are unaltered β ∝ λ

∴ $\frac{β'}{β} = \frac{λ'}{λ}$

β' = β × $\frac{λ'}{λ}$

= 1 × $\frac{6000}{1600}$

⇒ β = 1.2 mm

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