The Rayleigh Criterion and the Wavelength of Light

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(b) A car passes you on the highway and you notice the taillights of the car are 1.16 m apart. Assume that the pupils of your eyes have a diameter of 7.2 mm and index of refraction of 1.36. Given that the car is 14.1 km away when the taillights appear to merge into a single spot of light because of the effects of diffraction, what wavelength of light does the car emit from its taillights (what would the wavelength be in vacuum)?


Okay so i know the formula is

Theta = Index of Refraction* Wavelength / Distance

How do i find the limiting angle of resolution in order to get the wavelength
 
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Alt+F4 said:
(b) A car passes you on the highway and you notice the taillights of the car are 1.16 m apart. Assume that the pupils of your eyes have a diameter of 7.2 mm and index of refraction of 1.36. Given that the car is 14.1 km away when the taillights appear to merge into a single spot of light because of the effects of diffraction, what wavelength of light does the car emit from its taillights (what would the wavelength be in vacuum)?


Okay so i know the formula is

Theta = Index of Refraction* Wavelength / Distance

How do i find the limiting angle of resolution in order to get the wavelength
You want to be using the "Rayleigh Criterion". It's the one with the factor of 1.22 in it.
 
Chi Meson said:
You want to be using the "Rayleigh Criterion". It's the one with the factor of 1.22 in it.
ok but then Distance would be 13.5 Km and what is the limiting angle?

(1.22)(X) / (7.2/1000) = ?