Run 2 17. Set # of slits to 2 18. Set Wave Length to 400nm 19. Set Slit width to 1600 nm 20. Set Slit spacing to 5000nm In row 18 21. Record distance to 1st bright fringe 22. Record distance to 2nd bright fringe 23. Record distance to 3rd bright fringe Knowing the screen distance to be 1m 24. Calculated the measured angle to 1st bright fringe 25. Calculated the measured angle to 2nd bright fringe 26. Calculated the measured angle to 3rd bright fringe Using sin(θ)=mλ/d 27. Calculate θ for 1st bright fringe

a) m=1, y₁ = 0.08 m , θ₁ = 4.57º ,  b) m=2,  y₂ = 0.16 m , θ₂ = 9.09º ,  c)  m=3,   y₃ = 0.24 m ,   θ₃ = 13.5º

Explanation:

After reading your strange statement, I understand that this is an interference problem, I transcribe the data to have it more clearly. Number of slits 2, distance between slits 5000 nm, wavelength 400 nm, distance to the screen 1 m.

They ask us to calculate the angles for the first, second and third interference, they also ask us to write down the distance from the central maximum.

The expression for constructive interference for two slits is

             d sin θ = m λ

where d is the distance between the slits, λ is the wavelength used, m is an integer representing the order of interference

Let's use trigonometry to find the distance from the central maximum

         tan θ = y / L

in all interference experiments the angle is small,

         tan θ = sin θ / cos θ = sin θ

         sint θ = y / L

let's replace

        d y / L = m λ

        y = m λ L / d

let's calculate

distance to the first maximum m = 1

          y₁ = 1  400 10⁻⁹ 1/5000 10⁻⁹

          y₁ = 0.08 m

distance to second maximum m = 2

          y₂ = 2  400 10⁻⁹ 1/5000 10⁻⁹

          y₂ = 0.16 m

distance to the third maximum m = 3

          y₃ = 3  400 10⁻⁹ 1/5000 10⁻⁹

          y₃ = 0.24 m

with these values we can search for each angle

           tan θ = y / L

           θ = tan⁻¹ y / L

for m = 1

            θ₁ = tan⁻¹ (0.08 / 1)

            θ₁ = 4.57º

for m = 2

            θ₂ = tan⁻¹ (0.16 / 1)

            θ₂ = 9.09º

for m = 3

            θ₃ = tan⁻¹ (0.24 / 1)

            θ₃ = 13.5º

c. convergent boundary.

explanation:

50 inches should be the resultant vector