First slide

Young's double slit Experiment

Question

In a Young's double slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two wavelength $λ_{0} = 750 nm and λ = 900 nm$ . 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

Moderate

A

1.5 mm
B

3 mm
C

4.5 mm
D

6 mm

Solution

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)^th order fringe for $λ_{2} .$
Therefore $\frac{{nλ}_{1} D}{d} = \frac{(n + 1) λ_{2} D}{d}$
$\Rightarrow n \times 900 \times 10^{- 9} = (n + 1) 750 \times 10^{- 9} \Rightarrow n = 5$
Minimum distance from central maxima
$= \frac{{nλ}_{1} D}{d} = \frac{5 \times 900 \times 10^{- 9} \times 2}{2 \times 10^{- 3}} = 4 .5 mm$

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