Light from two lasers is incident on an opaque barrier with a single slit of width 4.0 x 10-5 m. One laser emits light of wavelength 4.0 x 10-7 m and the other is 6.0 x 10-7 m. A screen to view the light intensity pattern is 2.0 m behind the barrier. What is the distance from the center of the pattern to the nearest completely dark spot (dark for both colors)?

y = 6 10⁻² m

Explanation:

This is a diffraction exercise which is described by the expression

          a sin θ = m λ

we can use trigonometry to find the distance from the inside of the screen to the dark point (y)

          tan θ = y / L

angles are very small in diffraction experiments, so we can approximate

          tan θ = sin θ /cos θ = sin θ

          sin θ = y / L

substituting

         a (y / L) = m λ

Let's find the points where the intensity becomes zero

         y = m L λ / a

λ₁ = 4.0 10⁻⁷ m

m = 1

          y = 1 2.0 4.0 10⁻⁷/ 4.0 10⁻⁵

          y = 2 10⁻² m

m = 2

          y = 4 10⁻² m

λ₂ = 6.0 10⁻⁷ m

m = 1

          y = 1 2.0 6.0 10⁻⁷ / 4.0 10⁻⁵

          y = 3 10⁻² m

m = 2

          y = 6 10⁻² m

if we want a point where the two colors are dark, we set the two expressions equal

          y₁ = y₂

          m₁ L  λ₁ / a = m₂ L  λ₂ / a

          m₁/m₂ =  λ₂/λ₁

          m₁ / m₂ = 6 10⁻⁷ / 4 10⁻⁷

          m₁ / m₂ = 1.5

since the quantities m must be integers, the smallest relation that the relation fulfills is

            m₁ = 3

            m₂ = 2

the distance for this destructive interference is

           y = 3 2 4.0 10⁻⁷ / 4.0 10⁻⁵

           y = 6 10⁻² m

this is the first point where the minimum of the two wavelengths coincide

well, you would divide 135 by 33 to get your answer. then that is how many seconds it would take. boardman should fly about 132 m or 4.090909090909091 seconds.