Physics, 12.08.2020 08:01 july00

Coherent light with wavelength 601 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringe. For what wavelength of light will thefirst-order dark fringe be observed at this same point on the screen?

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Answer from: gibbss80stu

The wavelength is $\lambda = 1805 nm$

Explanation:

From the question we are told that

The wavelength of the light is $\lambda = 601 \ nm = 601 *10^{-9} \ m$

The distance of the screen is D = 3.0 m

The fringe width is $y = 4.84 \ mm = 4.84 *10^{-3} \ m$

Generally the fringe width for a bright fringe is mathematically represented as

$y = \frac{ \lambda * D }{d }$

=> $d = \frac{ \lambda * D }{ y }$

=> $d = \frac{ 601 *10^{-9} * 3}{ 4.84 *10^{-3 }}$

=> $d = 0.000373 \ m$

Generally the fringe width for a dark fringe is mathematically represented as

$y_d = [m + \frac{1}{2} ] * \frac{\lambda D }{d }$

Here m = 0 for first order dark fringe

$y_d = [0 + \frac{1}{2} ] * \frac{\lambda D }{d }$

looking at which we see that y_d = y

$4.84 *10^{-3} = [0 + \frac{1}{2} ] * \frac{\lambda * 3 }{ 0.000373 }$

=> $\lambda = 1805 *10^{-9} \ m$

=> $\lambda = 1805 nm$

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