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

Answer:

$3.38\cdot 10^9 m$

Explanation:

The formula for the double-slit interference pattern is:

$y=\frac{m\lambda D}{d}$

where:

m is the order of the maximum

$\lambda$ is the wavelength

D is the distance between the slits and the screen

d is the distance between the two slits

If we assume for instance m=5, the formula gives us the distance between the fifth maximum and the first maximum. However, this will also gives us the distance between the fifth minimum and the first minimum, as the minima fall exactly between two maxima.

Therefore, if we use:

$m=5\\d=0.22 mm=2.2\cdot 10^{-4} m\\y=59 cm=0.59 m\\D=6.3 mm=6.3\cdot 10^{-3}m$

We can find the wavelength of the light:

$\lambda=\frac{yD}{md}=\frac{(0.59 m)(6.3\cdot 10^{-3} m)}{(5)(2.2\cdot 10^{-4} m)}=3.38 m=3.38\cdot 10^9 m$

Answer 2

Distance between two slits is given as

$d = 0.22 mm = 0.22 \times 10^{-3} m$

distance of screen and slits is given as

$L = 59 cm$

now the position of minimum intensity on the screen given as

$y = \frac{(2N-1)\lambda L}{2d}$

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

$y_5 - y_1 = \frac{(9 - 1)\lambda L}{2d}$

now plug in all values

$6.3 \times 10^{-3} = \frac{4 \times \lambda\times 0.59}{0.22 \times 10^{-3}}$

$\lambda = 587.3 nm$

so above is the wavelength of light