A double-slit interference pattern is created by two narrow slits spaced 0.21 mm apart. The distance between the first and the fifth minimum on a screen 59 cm behind the slits is 5.9 mm.What is the wavelength of the light used in this experiment?

Respuesta :

Answer:

[tex]5.25\times10^{-8} m[/tex]

Explanation:

[tex]d[/tex] = separation of the slits = 0.21 mm = 0.00021 m

[tex]D[/tex] = Screen distance = 59 cm = 0.59 m

[tex]\lambda[/tex] = wavelength of the light

tex]y_{n}[/tex] = location of nth minima on the screen

[tex]y_{5}[/tex] = location of fifth minima on the screen

[tex]y_{1}[/tex] = location of first minima on the screen

location of nth minima on the screen is given as

[tex]y_{n} = \frac{(2n - 1) D \lambda}{2d}[/tex]

For n = 1

[tex]y_{1} = \frac{(2(1) - 1) D \lambda}{2d}\\y_{1} = \frac{(0.5) D \lambda}{d}[/tex]

For n = 5

[tex]y_{5} = \frac{(2(5) - 1) D \lambda}{2d}\\y_{1} = \frac{(4.5) D \lambda}{d}[/tex]

Given that:

[tex]y_{5} - y_{1} = 0.00059\\\frac{(4.5) D \lambda}{d} - \frac{(0.5) D \lambda}{d} = 0.00059\\\frac{(4) D \lambda}{d} = 0.00059\\\frac{(4) (0.59) \lambda}{0.00021} = 0.00059\\ \lambda = 5.25\times10^{-8} m[/tex]

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