Single-Slit Diffraction Peak Width

Classical Mechanics Level 2

In a single-slit diffraction experiment with slit width $a$ , distance $d$ between barrier and measurement screen, and light of wavelength $\lambda$ , what is the width of the central maximum peak?

\frac{\lambda a}{d}

\frac{2\lambda d}{a}

\frac{2\lambda a}{d}

\frac{\lambda d}{a}

1 solution

Matt DeCross
Mar 13, 2016

The width of the central maximum peak is given by the distance between the two adjacent minima. Since the minima are located at:

$y = \frac{m \lambda d }{a},$

for $m$ a nonzero integer, the width is:

$\Delta y = \frac{\lambda d}{a} - - \frac{\lambda d}{a} = \frac{2\lambda d}{a}.$

Single-Slit Diffraction Peak Width

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