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A Young's double-slit interference experiment is performed with monochromatic light. The separation between the slits is 0.48 mm. The interference pattern on the screen 3.7 m away shows the first maximum 5.1 mm from the center of the pattern. What is the wavelength of the light in nm

Answer :

Answer:

Wavelength of light used will be equal to 661 nm

Explanation:

We have given distance between slits d = 0.48 mm = [tex]0.48\times 10^{-3}m[/tex]

It is given interference pattern is 3.7 m away so D = 3.7 m

First maximum from the center is 5.1 mm

So [tex]y=5.1\times 10^{-3}m[/tex]

Distance of the first maximum from the center is equal to [tex]y=\frac{\lambda D}{d}[/tex]

[tex]5.1\times 10^{-3}=\frac{\lambda\times 3.7}{0.48\times 10^{-3}}=661\times 10^{-9}=661nm[/tex]

So wavelength of light used will be equal to 661 nm

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