Physics, 24.03.2020 20:07 ameliaduxha7

Coherent light with wavelength 608 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 thescreen?
Express your answer in micrometers(not in nanometers).

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Answer from: wjdelrieovie5s

$1.22 \mu m$

Explanation:

In the double-slit interference, light passes through a double slit and produce a pattern of alternating bright and dark fringes on a distant screen. This pattern is due to the combined effect of the diffraction of each slit + the interference of the light coming from the two slits.

The condition to observe a maximum (bright fringe), so costructive interference, in the distant screen, is:

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

where:

y is the distance of the m-th maximum from the central fringe

$\lambda$ is the wavelength of the light used

D is the distance of the screen from the slits

d is the separation between the slits

In this problem, we know that:

$\lambda=608 nm=608\cdot 10^{-9}m$ is the wavelength of light used

D=3.00 m is the distance of the screen

$y=4.84 mm = 4.84\cdot 10^{-3} m$ is the distance of the first maximum (first-order bright fringe) from the central pattern, so when

m = 1

Solving for d, we find the separation of the slits:

$d=\frac{m\lambda D}{y}=\frac{(1)(608\cdot 10^{-9})(3.00)}{4.84\cdot 10^{-3}}=3.77\cdot 10^{-4} m$

The first dark fringe on the screen instead is given by the formula

$y'=\frac{(\frac{\lambda'}{2})D}{d}$

where

$\lambda'$ is the wavelength of the new light

Here we want the first dark fringe of the new light to be coincident to the first bright fringe of the previous light, so

$y=4.84\cdot 10^{-3}m$

Therefore, solving for $\lambda'$ ,

$\lambda'=\frac{2y'd}{D}=\frac{2(4.84\cdot 10^{-3})(3.77\cdot 10^{-4})}{3.00}=1.22\cdot 10^{-6} m = 1.22 \mu m$

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