<p>The (weird for historical reasons) defintion of magnitude is that a difference of 5 magnitudes corresponds to a factor of 100 in the brightness of the source.</p>

<p>So a difference of 3-(-27) = 30 magnitudes is a difference of $10^{12}$ in brightness. </p>

<p>The surface brightness of the mirror and the sun is effectively equal, since the mirror reflects 100% of the sun's light, so the only reason for the sun to be brighter is that it is larger. The brightness is in direct proportion to the apparent area of the source.</p>

<p>This is because brightness is a simple linear scale. Two suns of the same size and brightness would be twice as bright as one. So if the sun is $10^{12}$ times brighter it must cover an area $10^{12}$ times larger.</p>



CC BY-SA 4.0
2018-08-15T12:18:11.197Z
Could you show me the formulas used to derive this conclusion - my understanding of flux density does not lead me to conclude this, even though it makes intuitive sense?
Tosic
CC BY-SA 4.0
2018-08-15T12:49:27.193Z
There's no real formula.  If you have two sources of illuminance a lux and b lux, then the total illuminance is (a+b) lux. If you double the (apparent) area of a source, without changing the surface brightness, you double the illumination. If the use of lux feels confusing, then just think of "watts per square metre". The power of two suns is double the power of one sun. The illumination of two suns is double that one sun. The area covered by a source 10^12 times brighter is 10^12 times larger.  Simple linearity.  L = kA
James K
CC BY-SA 4.0
2018-08-15T13:54:18.577Z
I started by wonderng:   Brightness is a function of total exitance from a surface, but that total from the sun is very close to a point source from the point of view of the moon, so the illuminance at the moon varies as the square of the distance from the sun.  The mirror won't have the sun's surface brightness unless you use a lens to re-image the sun. **edit** upon reflection, your answer appears to be consistent with that comment.
Carl Witthoft
CC BY-SA 4.0
2018-08-16T09:36:51.243Z
@CarlWitthoft How is it consistent? How can the surface brightness be the same when it equals the flux density on the surface divided by pi, and if the flux density on the surface were equal, would that not mean that the mirror somehow "grabbed" all the light from the Sun?
Tosic
CC BY-SA 4.0
2018-08-16T11:33:07.950Z
You are over complicating.  Consider an even and isotropic light source eg a light box. The surface brightness of the light box does not depend on distance. The total brightness is proportional to the apparent area of the box.  A mirror is placed that reflects the light box. The surface brightness of mirror is equal to that of the light box. The total brightness of the mirror is equal proportional to its apparent area.
James K
CC BY-SA 4.0
2018-08-16T13:10:52.547Z
All right, if I do make the surface brightness equal, I get that the square of the angular radius of the Sun is 10^12 that of the mirror as there extinction is considered negligible (and thus, opacity is 0 => thickness is 0 * r = 0), which is the correct answer ... I guess I'll just have to assume that a flat mirror does that, the problem is that other tasks from the same book feature a spherical mirror, whose surface brightness, according to the book, is not the same, could you elaborate on that (or should I post a separate question if this is too broad as a comment)?
Tosic
CC BY-SA 4.0
2018-08-15T09:03:38.070Z
2018-08-15T09:03:38.070Z
James K
<p>I was solving a task that said (paraphrasing):</p>

<blockquote>
  <p>There is a flat mirror on the Moon, whose reflection coefficient is 100%, and observers on the Earth observe it as a star whose apparent magnitude is $3^m$. Knowing the magnitude of the Sun is $-27^m$, and its angular diameter is $0.5 {^\circ}$, calculate the diameter of the mirror.</p>
</blockquote>

<p>The solution used Pogson's formula to state that the Sun is $1e12$ times brighter than the mirror ... and then the author said it is "obvious" that implies the surface of the mirror's disk in the sky, in $({^\circ})^2$, is $1e-12$ times that of the Sun.</p>

<p>How is it obvious, how do we derive this relation?</p>

<the-sun><the-moon><apparent-magnitude><angular-diameter><mirror>
How would a flat mirror on the Moon reflect sunlight
Tosic
# How would a flat mirror on the Moon reflect sunlight

(Asked by: username_0 on 2018-08-15)

I was solving a task that said (paraphrasing):

> There is a flat mirror on the Moon, whose reflection coefficient is 100%, and observers on the Earth observe it as a star whose apparent magnitude is $3^m$. Knowing the magnitude of the Sun is $-27^m$, and its angular diameter is $0.5 {^\circ}$, calculate the diameter of the mirror.

The solution used Pogson's formula to state that the Sun is $1e12$ times brighter than the mirror ... and then the author said it is "obvious" that implies the surface of the mirror's disk in the sky, in $({^\circ})^2$, is $1e-12$ times that of the Sun.

How is it obvious, how do we derive this relation?

**Tags:** the-sun, the-moon, apparent-magnitude, angular-diameter, mirror

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## 1 answers

### Answer by username_1 on 2018-08-15. Score: 4 | Accepted answer

The (weird for historical reasons) defintion of magnitude is that a difference of 5 magnitudes corresponds to a factor of 100 in the brightness of the source.

So a difference of 3-(-27) = 30 magnitudes is a difference of $10^{12}$ in brightness.

The surface brightness of the mirror and the sun is effectively equal, since the mirror reflects 100% of the sun's light, so the only reason for the sun to be brighter is that it is larger. The brightness is in direct proportion to the apparent area of the source.

This is because brightness is a simple linear scale. Two suns of the same size and brightness would be twice as bright as one. So if the sun is $10^{12}$ times brighter it must cover an area $10^{12}$ times larger.

> **Comments:**
> **username_0**: Could you show me the formulas used to derive this conclusion - my understanding of flux density does not lead me to conclude this, even though it makes intuitive sense?  
> **username_1**: There's no real formula.  If you have two sources of illuminance a lux and b lux, then the total illuminance is (a+b) lux. If you double the (apparent) area of a source, without changing the surface brightness, you double the illumination. If the use of lux feels confusing, then just think of "watts per square metre". The power of two suns is double the power of one sun. The illumination of two suns is double that one sun. The area covered by a source 10^12 times brighter is 10^12 times larger.  Simple linearity.  L = kA  
> **username_2**: I started by wonderng:   Brightness is a function of total exitance from a surface, but that total from the sun is very close to a point source from the point of view of the moon, so the illuminance at the moon varies as the square of the distance from the sun.  The mirror won't have the sun's surface brightness unless you use a lens to re-image the sun. **edit** upon reflection, your answer appears to be consistent with that comment.  
> **username_0**: @CarlWitthoft How is it consistent? How can the surface brightness be the same when it equals the flux density on the surface divided by pi, and if the flux density on the surface were equal, would that not mean that the mirror somehow "grabbed" all the light from the Sun?  
> **username_1**: You are over complicating.  Consider an even and isotropic light source eg a light box. The surface brightness of the light box does not depend on distance. The total brightness is proportional to the apparent area of the box.  A mirror is placed that reflects the light box. The surface brightness of mirror is equal to that of the light box. The total brightness of the mirror is equal proportional to its apparent area.  
> **username_0**: All right, if I do make the surface brightness equal, I get that the square of the angular radius of the Sun is 10^12 that of the mirror as there extinction is considered negligible (and thus, opacity is 0 => thickness is 0 * r = 0), which is the correct answer ... I guess I'll just have to assume that a flat mirror does that, the problem is that other tasks from the same book feature a spherical mirror, whose surface brightness, according to the book, is not the same, could you elaborate on that (or should I post a separate question if this is too broad as a comment)?  

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