Question 24.6: A Single-Slit Experiment Goal Find the positions of the dark...

A Single-Slit Experiment

Goal Find the positions of the dark fringes in single-slit diffraction.

Problem Light of wavelength 5.80 \times 10^{2} \mathrm{~nm} is incident on a slit of width 0.300 \mathrm{~mm}. The observing screen is placed 2.00 \mathrm{~m} from the slit. Find the positions of the first dark fringes and the width of the central bright fringe.

Strategy This problem requires substitution into Equation 24.11

\sin\,\theta_{\mathrm{dark}}\,=\,m\,{\frac{\lambda}{a}}\,\qquad m=\,\pm\,1,\,\pm\,2,\,\pm\,3,\,…         (24.11)

to find the sines of the angles of the first dark fringes. The positions can then be found with the tangent function, since for small angles \sin \theta \approx \tan \theta. The extent of the central maximum is defined by these two dark fringes.

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The first dark fringes that flank the central bright fringe correspond to m= \pm 1 in Equation 24.11:

\sin \theta= \pm \frac{\lambda}{a}= \pm \frac{5.80 \times 10^{-7} \mathrm{~m}}{0.300 \times 10^{-3} \mathrm{~m}}= \pm 1.93 \times 10^{-3}

Use the triangle in Figure 24.18 to relate the position of the fringe to the tangent function:

\tan \theta=\frac{y_{1}}{L}

Because \theta is very small, we can use the approximation \sin \theta \approx \tan \theta and then solve for y_{1} :

\begin{aligned} \sin \theta & \approx \tan \theta \approx \frac{y_{1}}{L} \\ y_{1} & \approx L \sin \theta= \pm L \frac{\lambda}{a}= \pm 3.86 \times 10^{-3} \mathrm{~m} \end{aligned}

Compute the distance between the positive and negative first-order maxima, which is the width w of the central maximum:

w=+3.86 \times 10^{-3} \mathrm{~m}-\left(-3.86 \times 10^{-3} \mathrm{~m}\right)=7.72 \times 10^{-3} \mathrm{~m}

Remarks Note that this value of w is much greater than the width of the slit. However, as the width of the slit is increased, the diffraction pattern narrows, corresponding to smaller values of \theta. In fact, for large values of a, the maxima and minima are so closely spaced that the only observable pattern is a large central bright area resembling the geometric image of the slit. Because the width of the geometric image increases as the slit width increases, the narrowest image occurs when the geometric and diffraction widths are equal.

24.18

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