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Coherent light, with wavelength of 594nm, passes through two very narrow slits, and the...

Question:

Coherent light, with wavelength of 594nm, passes through two very narrow slits, and the interference pattern is observed on a screen, a distance of 3.00m from the slits. The first-order bright fringe is a distance of 4.84mm from the center of the central bright fringe. For what wavelength of light will the first-order dark fringe be observed at this same point on the screen in micrometers?

Young's double slit experiment

In Young's double slit experiment, the angular position ({eq}\theta {/eq}) of the minimum and maximum interference can be determined by the expression:

{eq}d\sin\theta=\left\{\begin{matrix} m\lambda\,\,\,\,\,\,\,\max\\ \left (m+\frac{1}{2} \right )\lambda\,\,\,\,\,\,\,\min \end{matrix}\right. {/eq}

where {eq}d {/eq} is the separation between the slits, {eq}m {/eq} is the order and {eq}\lambda {/eq} the wavelength used.

In addition, separation of any point of the interference pattern from the center of the pattern ({eq}y {/eq}) can be calculated using the trigonometric relationship: {eq}\tan\theta=\frac{y}{D} {/eq}

where {eq}D {/eq} is the distance at which the interference pattern is observed.

Answer and Explanation:

If we consider the approximation of small angle, and combine the equations the angular position with the equation of the distance to the center of the interference pattern we obtain that:

{eq}\sin\theta\sim \tan\theta\Rightarrow d\sin\theta=d\left (\frac{y}{D} \right )=\left\{\begin{matrix} m\lambda\,\,\,\,\,\,\,\max\\ \left (m+\frac{1}{2} \right )\lambda\,\,\,\,\,\,\,\min \end{matrix}\right.\\ \therefore y_m=\left\{\begin{matrix} \frac{mD\lambda}{d}\,\,\,\,\,\,\,\max\\ \left (m+\frac{1}{2} \right )\frac{D\lambda}{d}\,\,\,\,\,\,\,\min \end{matrix}\right. {/eq}

For the maximum of first order, it is fulfilled that:

{eq}\left.\begin{matrix} m=1\\ y_m(\max)=\frac{mD\lambda}{d} \end{matrix}\right\}\Rightarrow y_1(\max)=\frac{D\lambda_1}{d} {/eq}

Using this result and the expression for the minimum order zero is obtained:

{eq}\left.\begin{matrix} m=0\\ y_m(\min)=\left ( m+\frac{1}{2} \right )\frac{D\lambda_0}{d} \end{matrix}\right\}\Rightarrow y_0(\min)=\frac{D\lambda_0}{2d}\\ \left.\begin{matrix} y_1(\max)=\frac{D\lambda_1}{d}\\ y_0(\min)=\frac{D\lambda_0}{2d} \end{matrix}\right\}\Rightarrow y_1(\max)=y_0(\min)\Rightarrow \frac{D\lambda_1}{d}=\frac{D\lambda_0}{2d}\Rightarrow \lambda_0=2\lambda_1\\ \therefore \lambda_0=2\mathrm{(0.594\,\mu m)=1.19\,\mu m} {/eq}

That is, for the first minimum to occupy the position of the first order maximum, it is necessary to double the wavelength value.


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Constructive and Destructive Interference

from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 8 / Lesson 16
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