Question

manufacturers of wire sometimes use a laser to continually monitor the thickness of the product. the wire intercepts the laser beam, producing a different diffraction pattern like that of a single slit of the same width as the wire diameter. suppose a laser of wavelength 649.2 nm illuminates a wire, and the diffraction pattern appears on a screen 2.92 m away. if the desired wire diameter is 1.45 mm, what is the observed distance between the two tenth order minima

. do not use the numbers given in diagram.

Answer 1

The equation needed is the following:

$y=\frac{m\lambda L}{d}$

here m is equal to 10

wavelength is equal to 649.2 nm

L is equal to 2.92m

d is equal to 1.45 mm

so:

$y=\frac{10(649.2X10^{-9}) 2.92}{1.45X10^{-3}}\rightarrow y=0.01307m$

multiply this value by two because it is the distance between the 2 tenth order minima:

Answer

$\boldsymbol{y=2.614\: cm}$