## Coherent light with wavelength 601 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance

Question

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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1 year 2021-08-09T00:07:57+00:00 1 Answers 122 views 0

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

So

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

looking at which we see that   $$y_d = y$$

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

=>    $$\lambda = 1805 *10^{-9} \ m$$

=>    $$\lambda = 1805 nm$$