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

it's a b and

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr