In a Young's double slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two wavelength λ0=750 nm and λ=900 nm. The minimum distance from the common central bright fringe on a screen 2 m from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is

Question

In a Young's double slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two wavelength λ0=750 nm and λ=900 nm. The minimum distance from the common central bright fringe on a screen 2 m from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is (A) 1.5 mm (B) 3 mm (C) 4.5 mm (D) 6 mm

Tardigrade Admin · Accepted Answer

From the given data, note that the fringe width

(β_{1})

for

λ_{1} = 900 nm

is greater than fringe width

(β_{2})

for

λ_{2} =

750 nm

. This means that at though the central maxima of the two coincide, but first maximum for

λ_{1} = 900

nm

will be further away from the first maxima for

λ_{2} = 750 nm

, and so on. A stage may come when this mismatch equals

β_{2}

, then again maxima of

λ_{1} = 900

nm

, will coincide with a maxima of

λ_{2} = 750 nm

let this correspond to nth order fringe for

λ_{1}

. Then it will correspond to

(n + 1)^{t h}

order fringe for

λ_{2}

.
Therefore

\frac{n λ _{1} D}{d} = \frac{( n + 1 ) λ _{2} D}{d}

\Rightarrow n \times 900 \times 1 0^{- 9} = (n + 1) 750 \times 1 0^{- 9} \Rightarrow n = 5

Minimum distance from central maxima

= \frac{n λ _{1} D}{d} = \frac{5 \times 900 \times 1 0 ^{- 9} \times 2}{2 \times 1 0 ^{- 3}} = 4.5 mm