Question

A double-slit interference pattern is created by two narrow slits spaced 0.22 mm apart. The distance between the first and the fifth minimum on a screen 59 cm behind the slits is 6.3 mm . You may want to review (Pages 629 - 634) . For help with math skills, you may want to review: Rearrangement of Equations Involving Multiplication and Division Conversion Factors What is the wavelength (in nm) of the light used in this experiment?

Answer 1

Distance between two slits is given as

`d = 0.22 mm = 0.22 * 10^(-3) m`

distance of screen and slits is given as

`L = 59 cm`

now the position of minimum intensity on the screen given as

`y = ((2N-1)\lambda L)/(2d)`

now for the distance between fifth minima and first minima we can say

`y_5 - y_1 = ((9 - 1)\lambda L)/(2d)`

now plug in all values

`6.3 * 10^(-3) = (4 * \lambda* 0.59)/(0.22 * 10^(-3))`

`\lambda = 587.3 nm`

so above is the wavelength of light