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.

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.