How will light react in this situation....

In summary, the individual is seeking help with calculating shadows on a flat surface when the light source is at a 0 or 180 degree angle. They mention their knowledge of light traveling in straight lines and the difficulty in finding the intersection of the shadow volume with the ground. They also mention the possibility of using GPUs for these types of calculations. The conversation touches on the concept of a point source versus an extended source and the potential difficulty in achieving a 0 or 180 degree angle in nature.
  • #1
xrayspecs
2
2
1595536479244.png


1595539199288.png

Dear Physics Forum,

I need help with this problem. In the diagrams above I try to show my difficulty. My main problem is working out how shadow will fall on a completely flat surface with the light source at more or less a 0 or 180 degree angle (depending on how you want to look at it) . Through internet searches I have seen how it is possible to calculate the shadow in a natural setting at around 45 degrees.

From knowledge acquired at school and light research... (pun intended) I know light travels in straight lines. So if it met a completely flat surface how would it be possible to work out where the shadow of the shape would appear on the ground? And how long it would measure on said ground?

Please let me know if you need any more information or drawings to try and make this situation easier to understand.
Any help will be most appreciated!

Thank you for reading,

Xrayspecs
 
  • Like
Likes Dale
Science news on Phys.org
  • #2
xrayspecs said:
And how long it would measure on said ground?
Infinite, if the ground is an infinite flat plane.
 
  • #3
xrayspecs said:
I know light travels in straight lines.

Yes, but the problem is that different straight lines don't necessarily have the same direction.

xrayspecs said:
So if it met a completely flat surface how would it be possible to work out where the shadow of the shape would appear on the ground?

In general you need to find the intersection of the shadow volume with the ground. In case of a sphere and a point source the shadow volume is a cone and the intersection with the ground is an ellipse, parabola or hyperbola. In case of real bodies and light sources it will me much more complex. This kind of calculations is what GPUs are build for.

xrayspecs said:
And how long it would measure on said ground?

As mentioned above this can be quite difficult to answer, especially if the shape of the object is not clearly defined. But you can at least estimate the maximum length of the umbra. It is the distance where the object has the same angular diameter as the light source. In sunlight that would be 106 to 109 times the size of the object.
 
  • #4
You have to understand that the picture you drew is misleading. The Sun is so far away that its rays are parallel at any given location. A modified picture is shown below. What do you think the shadow would be like on the flat plane? Remember that shadow is the absence of light. In the picture below, what light, which would normally fall on the plane, is blocked by the object?

Shadow.png
 
Last edited:
  • Like
Likes berkeman
  • #5
If we're treating the sun as a point source (which it isn't), then the shadow would lie between the two thin lines in the OP -- the ones that eminate from the lower-left and lower-right "corners" of the object. And as A.T. said, it would extend to infinity.

If we're treating the sun as an extended source, then each of those lines becomes a pair of lines, at roughly 0.5 degrees to each other. In between the two lines, there is a transition from fully bright to fully dark. But the shadow still extends to infinity.

The answer to the question "How would I show this curve?", is that you would not show it at all.
 
  • #6
xrayspecs said:
... how shadow will fall on a completely flat surface with the light source at more or less a 0 or 180 degree angle...
The problem I see is the difficulty to achieve that angle in nature.
Even before the Sun rises, some light hits our eyes and objects.
That light comes from reflection on dust particles, bottom of clouds, etc.

If the object is tall enough, the shadow could be seen on the clouds.
Nd9GcTUKJ1-FfULViNKIkGQnVWM7BMXs2eDObBywA&usqp=CAU.jpg


moutain-cloud-shadow-1.jpg
 
  • Like
Likes PeroK
  • #7
If we are treating the sun as a point source then there is no shadow in the first place. The surface is not illuminated.
 
  • #8
Thank you for your comments!
 

1. How will light react when passing through a prism?

When light passes through a prism, it will refract or bend, separating into its component colors. This is due to the different wavelengths of light being bent at different angles as they pass through the prism.

2. How will light behave when reflecting off a smooth surface?

When light reflects off a smooth surface, it will follow the law of reflection, which states that the angle of incidence (incoming light) is equal to the angle of reflection (outgoing light). This results in a clear and undistorted reflection.

3. How will light react when traveling through a vacuum?

In a vacuum, light will travel at its maximum speed of approximately 299,792,458 meters per second. This is because there are no particles in a vacuum to slow down or obstruct the path of light.

4. How will light behave when passing through a medium with a higher refractive index?

When light passes through a medium with a higher refractive index, it will slow down and bend towards the normal (a line perpendicular to the surface of the medium). This is known as refraction and is responsible for phenomena such as the bending of light in a glass of water.

5. How will light react when encountering an object with a rough surface?

When light encounters an object with a rough surface, it will scatter in all directions due to the uneven surface. This results in a diffuse reflection, where the light is not reflected in a clear and organized manner, but rather in multiple directions.

Similar threads

Replies
3
Views
1K
Replies
22
Views
2K
Replies
40
Views
2K
Replies
12
Views
1K
Replies
7
Views
1K
Replies
6
Views
1K
Replies
5
Views
3K
  • Optics
Replies
12
Views
2K
  • Classical Physics
Replies
1
Views
2K
  • Materials and Chemical Engineering
Replies
12
Views
437
Back
Top