A Young’s interference experiment is performed with monochromatic light. The fir

Question

A Young’s interference experiment is performed with monochromatic light. The first maximum occurs 1.92 mm from the center of the interference pattern.What is the wavelength? Answer in units of nm.

The S1 and S2 on the left side have a distance between them of 0.665 mm. And the distance on the right is 1.92 mm just in case you can see it.

A Young?s interference experiment is performed with monochromatic light. The first maximum occurs 1.92 mm from the center of the interference pattern.What is the wavelength? Answer in units of nm. The S1 and S2 on the left side have a distance between them of 0.665 mm. And the distance on the right is 1.92 mm just in case you can see it.

Explanation / Answer

we have pat difference at a distance y from the center of the interference pattern is
= d*y/D where d=distance between the slits;D=distance of the screen
in this case path difference = (0.665/1000)*(1.92/1000)/3.12 = 4.09*10^-7 m
for the first maximum to occur the path difference =

so = 4.0923*10^-7 m = 409.23 nm

so,the wavelength is 409.23 nm

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