We know that
d sin(theta) = m lambda
m = 1
lambda = d sin(theta)
d = 10,000 per cm = 10,00000 lines per m
d = 1/1000000 = 10^-6 m
a)lambda1 = 10^-6 x sin24.1 = 408.33 x 10^-9 m
Hence, lambda1 = 408.33 nm
b)lambda2 = 10^-6 x sin30.5 = 507.54 x 10^-9 m
lambda2 = 507.54 nm
c)lambda3 = 10^-6 x sin36 = 587.79 x 10^-9 m
lambda2 = 587.79 nm
d)lambda4 = 10^-6 x sin41.6 = 663.93 x 10^-9 m
lambda2 = 663.93 nm
Parts a through d please. (17%) Problem is Passing an electric cu ent through an unknown...
An electric current through an unknown gas produces several distinct wavelengths of visible light. Consider the first order maxima for the wavelengths 401 nm, 429 nm, 525 nm, and 684 nm of this unknown spectrum, when projected with a diffraction grating of 5,000 lines per centimeter. a) What would the angle be (in degrees) for the 401 nm, 429 nm, 525 nm, and 684 nm lines. b)Using this grating, what would be the angle (in degrees) of the second-order maximum...