Question

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?

Answer 1

`5.25*10^(-8) m`

Step-by-step explanation:

`d` = separation of the slits = 0.21 mm = 0.00021 m

`D` = Screen distance = 59 cm = 0.59 m

`\lambda` = wavelength of the light

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

`y_(5)` = location of fifth minima on the screen

`y_(1)` = location of first minima on the screen

location of nth minima on the screen is given as

`y_(n) = ((2n - 1) D \lambda)/(2d)`

For n = 1

`y_(1) = ((2(1) - 1) D \lambda)/(2d)\\y_(1) = ((0.5) D \lambda)/(d)`

For n = 5

`y_(5) = ((2(5) - 1) D \lambda)/(2d)\\y_(1) = ((4.5) D \lambda)/(d)`

Given that:

`y_(5) - y_(1) = 0.00059\\((4.5) D \lambda)/(d) - ((0.5) D \lambda)/(d) = 0.00059\\((4) D \lambda)/(d) = 0.00059\\((4) (0.59) \lambda)/(0.00021) = 0.00059\\ \lambda = 5.25*10^(-8) m`