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In a Young's double-slit experiment, a set of parallel slits with a separation of 0.114 mm is illuminated by light having a wavelength of 558 nm and the interference pattern observed on a screen 4.50 m from the slits. (a) What is the difference in path lengths from the two slits to the location of a fourth order bright fringe on the screen? μm (b) What is the difference in path lengths from the two slits to the location of the fourth dark fringe on the screen, away from the center of the pattern?

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Answer:

a) 2232nm

b) 2511nm

Explanation:

a) To find the path difference Δl you use the following formula:

[tex]\Delta l=m\lambda[/tex]

m: order of a bright fringe

λ: wavelength of light = 558nm

for m=4:

[tex]\Delta l=(4)(558nm)=2232nm[/tex]

b) The path difference for the case of destructive interference you have:

[tex]\Delta l=(m+\frac{1}{2})\lambda[/tex]

m: order of a dark fringe

for m=4:

[tex]\Delta l=(4+\frac{1}{2})(558nm)=2511nm[/tex]

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