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

Answer:

$5.25\times10^{-8} m$

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

$y_{n} = \frac{(2n - 1) D \lambda}{2d}$

For n = 1

$y_{1} = \frac{(2(1) - 1) D \lambda}{2d}\\y_{1} = \frac{(0.5) D \lambda}{d}$

For n = 5

$y_{5} = \frac{(2(5) - 1) D \lambda}{2d}\\y_{1} = \frac{(4.5) D \lambda}{d}$

Given that:

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