Question

36.1 single-slit diffraction
in a single-slit diffraction experiment, a beam of monochromatic light of wavelength 508 nm is incident on a slit of width of 0.446 mm. if the distance between the slit and the screen is 2.50 m, what is the distance between the central axis and the first dark fringe (in mm)?
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Explanation:

Step1: Recall single - slit diffraction formula

For single - slit diffraction, the condition for dark fringes is \(a\sin\theta = m\lambda\), where \(a\) is the slit width, \(\theta\) is the angle of diffraction, \(m\) is the order of the dark fringe (\(m = 1,2,3,\cdots\)) and \(\lambda\) is the wavelength of light. For small angles (when the distance between the slit and the screen \(L\) is much larger than the distance \(y\) from the central axis to the fringe, i.e., \(L\gg y\)), \(\sin\theta\approx\tan\theta=\frac{y}{L}\). For the first dark fringe, \(m = 1\), so the formula for the distance \(y\) from the central axis to the first dark fringe is \(y=\frac{m\lambda L}{a}\).

Step2: Convert units

• Wavelength \(\lambda=508\space nm = 508\times10^{-9}\space m\)
• Slit width \(a = 0.446\space mm=0.446\times10^{-3}\space m\)
• Distance between slit and screen \(L = 2.50\space m\)
• Order of dark fringe \(m = 1\)

Step3: Substitute values into the formula

Substitute \(m = 1\), \(\lambda=508\times 10^{-9}\space m\), \(L = 2.50\space m\) and \(a=0.446\times 10^{-3}\space m\) into the formula \(y=\frac{m\lambda L}{a}\)

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Step4: Convert to millimeters

Since \(1\space m=1000\space mm\), then \(y = 2.84753\times 10^{-3}\space m\times1000\space mm/m=2.85\space mm\) (rounded to three significant figures)

Answer:

\(2.85\)