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A double-slit experiment is performed with light of wavelength 550 nm. The bright interference fringes are spaced 2.3 mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 360 nm?

Sagot :

okpalawalter8

Answer:

[tex]d_2=1.5*10^-3m[/tex]

Explanation:

From the question we are told that:

Initial Wavelength [tex]\lambda_1=550nm=550*10^{-9}[/tex]

Space 1 [tex]d_1=2.3*10^{-3}[/tex]

Final wavelength [tex]\lambda_2=360*10^{-9}[/tex]

Generally the equation for Fringe space at [tex]\lambda _2[/tex] is mathematically given by

 [tex]d_2=\frac{d_1}{\lambdaI_1}*\lambda_2[/tex]

 [tex]d_2=\frac{2.3*10^{-3}}{550*10^{-9}}*360*10^{-9}[/tex]

 [tex]d_2=1.5*10^-3m[/tex]

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