A Optical Limit of Maxwell’s Equations
631
A.5 Deflection of Sunlight
As an exercise in applying the ray equation, consider the image of the Sun as we
watch a sunset. That image is not in the direction of the actual Sun because of
refraction of sunlight in the atmosphere. 2
If we specify points by coordinates from the center of the Earth, then the index
of refraction will largely depend only on the distance r from the Earth center. Our
ray equation will now read
d
ds
n
dr
ds
=
dn
dr
r
which, when crossed with r, gives
nr ×
dr
ds
= constant ,
i.e.
nr sin θ = constant .
If the Sun is viewed on the horizon, θ R = 90 ◦ , so, along the ray,
r sin θ = Rn R .
We also know that the incident ray is a distance R + h from the center of the Earth,
where h is the effective height of the atmosphere, so we can replace r sin θ by R +
h. Finally, the angle cos between the incident ray and the refracted ray which
arrives at the Earth tangent to its surface is also the angle between the corresponding
normals, giving
cos =
R
R + h
=
1
n R
.
The value of the index of refraction at the Earth’s surface is 1.003, making = 4.4 ◦ .
Including the effect of light time delay, the Sun is seen 4.1 ◦ behind where it is
physically.
2 There is also an eight and a half minute time delay for light to get to our eyes from the Sun. The
Sun’s image moves so slowly across the sky that this effect is small, producing an angle deviation
of 8.5 min/24 × 60 min = 0.3 ◦ .
Précédent

- 643/703

Suivant