Characteristic orange light produced by sodium in a fl ame is due to an intense emission called the sodium D line, which is actually a doublet, with wavelengths (measured in vacuum) of 589.157 88 and 589.755 37 nm. The index of refraction of air at a wavelength near 589 nm is 1.000 292 6. Calculate the frequency, wavelength, and wavenumber of each component of the D line, measured in air.

a. i. 5.092 × 10²⁰ Hz ii. 588.98554 nm ii. 10667809.11 rad/m

b. i. 5.087 × 10²⁰ Hz ii. 589.58286 nm iii. 10657001.3 rad/m

Explanation:

refractive index, n = λ/λ'  where λ = wavelength in vacuum = and λ' = wavelength in air

a. For λ = 589.15788 nm,

i. Frequency,

f = c/λ where c = speed of light in vacuum = 3 × 10⁸ m/s and λ = 589.15788 nm = 589.15788 × 10⁻⁹ m

So, f = 3 × 10⁸ m/s ÷ 589.15788 × 10⁻⁹ m

= 0.005092 × 10¹⁷ /s

= 5.092 × 10²⁰ /s

= 5.092 × 10²⁰ Hz

ii. Wavelength,

Since n = λ/λ'  where λ = wavelength in vacuum = and λ' = wavelength in air

and n = 1.0002926

λ' = λ/n

= 589.15788 nm/1.0002926

= 588.98554 nm

k = 2π/λ'

= 2π/588.98554 nm

= 0.01066780911 rad/nm

= 0.01066780911 rad/nm × 10⁹ nm/1m

= 10667809.11 rad/m

b. For λ = 589.755 37 nm.,

i. Frequency,

f = c/λ where c = speed of light in vacuum = 3 × 10⁸ m/s and λ = 589.755 37 nm. = 589.755 37 × 10⁻⁹ m

So, f = 3 × 10⁸ m/s ÷ 589.755 37 × 10⁻⁹ m

= 0.005087 × 10¹⁷ /s

= 5.087 × 10²⁰ /s

= 5.087 × 10²⁰ Hz

ii. Wavelength,

Since n = λ/λ'  where λ = wavelength in vacuum = and λ' = wavelength in air

and n = 1.0002926

λ' = λ/n

= 589.755 37 nm./1.0002926

= 589.58286 nm

k = 2π/λ'

= 2π/589.58286 nm

= 0.0106570013 rad/nm

= 0.0106570013  rad/nm × 10⁹ nm/1m

= 10657001.3 rad/m

here's the link to your question!

explanation: