A single, monochromatic indigo light source is shined through an etched, flat prism with a slit separation of .0250mm. The resulting interference pattern is viewed on a screen 1.25m away. The third maximum is found to be 6.6cm from the central maximum. What is the wavelength of the indigo light

  λ = 440 nm

Explanation:

The phenomenon of constructive interference is described by the expressions

          d sin θ= m λ                    (1)

where d is the separation of the slits d = 0.0250 mm = 2.50 10⁻⁵ m, lam is the wavelength of the incident radiation and m is an integer indicating the order of interference

let's use trigonometry to find the angle

          tan θ = y / L

where L is the distance to the screen L = 1.25 m

in general interference experiments angles are very small

         tan θ =

         ten θ = sin θ

substituting

         sin θ = y / L

we substitute 1

          d y / L = m λ

          λ =

in the exercise indicate

        m = 3

         y = 6.6 cm = 6.6 10-2 m

we calculate

          λ = 2.50 10⁻⁵ 6.6 10⁻² /( 3   1.25)

          λ = 4.4 10⁻⁷ m

let's reduce to nm

          λ = 4.4 10⁻⁷ m (10⁹ nm / 1 m)

          λ = 440 nm

that would be b i just took the test : )

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.