Manufacturers of wire (and other objects of small dimensions) sometimes use a la

Question

Manufacturers of wire (and other objects of small dimensions) sometimes use a laser to continually monitor the thickness of the product. The wire intercepts the laser beam, producing a diffraction pattern like that of a single slit of the same width as the wire diameter (see Fig. below). Suppose a helium-neon laser, of wavelength 632.8 nm, illuminates a wire, and the diffraction pattern appears on a screen 2.60 m away. If the desired wire diameter is 0.18 mm, what would the observed width of the central maximumbe on the screen?

Explanation / Answer

Y = (m+1/2)R/d

Y = 1*632.8*10^-9*2.6/1.5*0.00018

Y = 4.5mm

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