Adouble-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?

Explanation:

= separation of the slits = 0.21 mm = 0.00021 m

= Screen distance = 59 cm = 0.59 m

= wavelength of the light

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

= location of fifth minima on the screen

= location of first minima on the screen

location of nth minima on the screen is given as

For n = 1

For n = 5

Given that:

4

explanation:

4

answer; i believe the correct answer is (40,000 j) good luck