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I'm trying to understand how smooth the reflecting ocean surface would need to be to produce such a small bright spot as seen from the DSCOVR satellite at Sun-Earth L1. It appears to be only about 8E-05 rad, or about 0.3 arcminutes.

My question is primarily about the geometrical optics involved in the reflection of light from the Sun off of the Earth. Interpretation of the apparent size and comparing to the image is secondary.

At a distance of 151.3 million km from Earth, the 1.391 million km diameter Sun has an angular width of 0.009196 radians. The DSCOVR satellite is 1.586 million km from the Earth looking at the Sun's reflection from somewhere around the Sun-Earth L1 point. It's close to the Earth-Sun axis (the Sun-Earth-DSCOVR angle is about 7.6 degrees) so let's simplify and assume it sits on-axis.

The Earth's radius is 6378 km; as a convex mirror its focal length is $f=-r/2$ or -3189 km.

Question: What would be the angular size of the Sun seen from DSCOVR reflecting in a spherical mirror model for the Earth in this configuration?


What I'm looking at (for context):

This sequence of images taken by the EPIC camera on the DSCOVR satellite near the Sun-Earth L1 point is notable for showing the progress of the Moon's shadow across the Earth, but that's not the topic of my question.

The first image shows the back-reflection of the Sun from the Earth. It is likely to be from a particularly calm spot on the Ocean, but that's not for sure, there are some images that show reflections from horizontally oriented ice crystals as well. See Are there measurements or calculations that suggest atmospheric ice plates would be horizontal to within 0.1 degrees? and also How could the recently explained “glints” seen by DSCOVR appear so compact considering the finite size of the sun? for some interesting images and more on that.

But that doesn't matter for the purposes of my question.

Width of Earth 6378 km x 2 / 1586000 km = 8.0428E-03 rad corresponds to ~1500 pixels.

Bright spot is about 15 pixels wide, or about 8E-05 rad.

Could such a small angular width be a geometrical reflection of the Sun from a very smooth ocean?

Raw image at NASA seen here.

enter image description here

enter image description here

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    $\begingroup$ Could you use the convex mirror formulas to calculate the size and location of the virtual image? $\endgroup$ – Keith McClary Mar 4 at 22:36
  • $\begingroup$ @KeithMcClary I expect one could but I'm not sure. For some reason the concept of virtual images always makes me dizzy. I think that the equations in your link $1/d_o + 1/d_i = 1/f$ and $m=-d_i/d_o$ might lead to a (de)magnification, but I am confused by the negative sign, and so I want to defer to the advice of some virtual experts on the topic ;-) $\endgroup$ – uhoh Mar 4 at 23:42
  • $\begingroup$ @KeithMcClary I've just added a bounty $\endgroup$ – uhoh Mar 11 at 4:05
  • $\begingroup$ astronomy.stackexchange.com/questions/29980/… $\endgroup$ – Muze the good Troll. Mar 13 at 19:23
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$d_o$ and $h_o$ are the distance and diameter of the Sun and $d_i$ and $h_i$ of the respective images. The Earth's radius is $r$ and $f=\frac{r}{2}$ is the focal length.

The mirror equation $$\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$$ in the approximation $d_o \gg f$ gives $d_i \approx f$. Then, using the magnification equation

$$h_i = \frac {d_i}{d_o}h_o \approx \frac {f}{d_o}h_o.$$

If $S$ is the distance to the satellite then the angular size is

$$\frac{h_i}{S} \approx \frac{h_o}{d_o}\frac{f}{S} = \frac{h_o}{d_o}\frac{r}{2S} \approx \frac{1.4\times10^6}{150\times10^6}\frac{6.4\times10^3}{2\times 1.6\times 10^6}\approx 1.8 \times 10^{-5}$$

where distances are in km. (We have also ignored that the image is $\frac{R}{2}$ closer to the satellite than is the Earth's center.)

On a more descriptive note, the reflection on perfectly smooth water would look like a very bright spot about 29 km in diameter. It is blurred out (due to sea waves) to about ten times that diameter on the image. It would be interesting to calculate the brightness (per area) of the (unblurred) spot compared to the brightness of the sunlit Earth's surface.

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  • $\begingroup$ @uhoh It would look like a very bright spot on the Earth 29 km in diameter. It is blurred out (due to sea waves) to about ten times that diameter on the image. It would be interesting to calculate the brightness (per area) of the (unblurred) spot compared to the brightness of the sunlit Earth's surface. $\endgroup$ – Keith McClary Mar 11 at 19:50
  • $\begingroup$ Great job, thanks for your help! $\endgroup$ – uhoh Mar 14 at 1:33
  • $\begingroup$ I've moved your comment up into the answer. It's quite informative and address the question directly so it's best to keep it in the post itself where future readers are more likely to read it. $\endgroup$ – uhoh Mar 14 at 1:34

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