Question

# In two-slit interference, if the slit separation is 14 $\mu$m and the slit widths are each 2.0 $\mu$m, how many two-slit maxima are in the central peak of the diffraction envelope?

Solution

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In the double slit experiment, there are two phenomena occurs, the first one is the interference due to the difference in the two paths, and the second one is the diffraction in the single slit. For the first minima in the diffraction pattern we substitute $m_1=1$, obtaining:

\begin{align}a\sin(\theta)&=\lambda\end{align}

The angular locations of the bright fringes of the double-slit interference pattern are given by:

\begin{align}d \sin(\theta)&=m_2 \lambda\end{align}

combine (1) and (2) to get:

$m_2=\dfrac{d}{a}$

where $d$ is the separation between the slits and $a$ is the width of the slit. Substitute with the givens to get:

\begin{align*}m_2&=\dfrac{14 \mathrm{~\mu m}}{2.0 \mathrm{~\mu m}} \\ &=7.0 \end{align*}

the seventh maximum missed, which means we only have 6 maximums in each side with a central maximum at $m_2=0$. Therefore, we have 13 maxima.

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