Adouble-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 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?

Explanation:

The formula for the double-slit interference pattern is:

where:

m is the order of the maximum

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:

We can find the wavelength of the light:

Distance between two slits is given as

distance of screen and slits is given as

now the position of minimum intensity on the screen given as

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

now plug in all values

so above is the wavelength of light

does anyone know or what?