A diffraction grating is placed 1.00 m from a viewing screen. Light from a hydrogen lamp goes through the grating. A hydrogen spectral line with a wavelength of 656 nm is seen 60.0 cm to one side of the center. Then, the hydrogen lamp is replaced with an unknown lamp. A spectral line is seen on the screen 36.4 cm away from the center. What is the wavelength of this spectral line

λ = 396.7 nm

Explanation:

For this exercise we use the diffraction ratio of a grating

           d sin θ = m λ

in general the networks works in the first order m = 1

we can use trigonometry, remembering that in diffraction experiments the angles are small

           tan θ = y / L

           tan θ = = sin θ

           sin θ = y / L

we substitute

          = m λ

with the initial data we look for the distance between the lines

           d =

           d = 1 656 10⁻⁹ 1.00 / 0.600

            d = 1.09 10⁻⁶ m

for the unknown lamp we look for the wavelength

           λ = d y / L m

           λ = 1.09 10⁻⁶ 0.364 / 1.00 1

           λ = 3.9676 10⁻⁷ m

           λ = 3.967 10⁻⁷ m

we reduce nm

           λ = 396.7 nm

i guess that visual aids present persuasive examples to what you are presenting, some factors would be theme (does it fit with the color, information, what you are presenting), size (can you see it, is it a big enough span of data/information), and overall look of the visual aid.

hope this : )

in order to follow the instructions and complete the statements, i need to use data from table c of your student guide. can you think of a way to let me see your student guide for a while ? i can't answer the question without the data in table c.