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In Young's double slit experiment a parallel light beam containing wavelength $λ_{1} = 4000 \overset{\circ}{A} and λ_{2} = 5600 \overset{\circ}{A}$ is incident on a diaphragm having two narrow slits. Separation between the slits is $d = 2 m m$ . If distance between diaphragm and screen is $D = 40 c m$ , calculate - distance between two consecutive black lines.

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$560 μ m$

$310 μ m$

$60 μ m$

$100 μ m$

answer is A.

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Detailed Solution

Since, interference pattern is always symmetric about central bright fringe, therefore, there are two first black lines one is at height $y$ from central bright fringe and the other at a depth $y$ from it.

Hence, distance between two consecutive black lines $= 2 y = 560 μ m$

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