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A light source, which emits two wavelength $λ_{1} =$ 400 nm and $λ_{2} =$ 600 nm, is used in a Young’s double slit experiment. If recorded fringe widths for $λ_{1}$ and $λ_{2}$ are $β_{1}$ and $β_{2}$ , and the number of fringes for them within a distance y on one side of the central maximum are m₁ and m₂ respectively, then

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$β_{2} > β_{1}$

$m_{1} > m_{2}$

From the central maximum, 3^rd maximum of $λ_{2}$ overlaps with 5^th minimum of $λ_{1}$

The angular separation of fringes for $λ_{1}$ is greater than $λ_{2}$

answer is A, B, C.

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Detailed Solution

We know that $β = \frac{λD}{d}$
As $λ_{2} > λ_{1}$

$∴ β_{2} > β_{1}$

$m = \frac{y}{β} \Rightarrow m \propto \frac{1}{β}$
Therefore, $m_{1} > m_{2}$

3rd maximum of $λ_{2}$ : $y_{2} = 3 \times \frac{λ_{2} D}{d} = 3 \times 600 \frac{D}{d} = 1800 \frac{D}{d}$

5th minimum of $λ_{1}$ : $y_{1} = \frac{(2 \times 5 - 1) λ_{1}}{2} \frac{D}{d} = 4.5 \times 400 \frac{D}{d} = 1800 \frac{D}{d}$
The angular width $= \frac{λ}{d}$

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