# Homework Help: Thought experiment with two point sources of light

1. Mar 30, 2012

You are in a dark room with no other sources of light
The walls are painted black, so no light is scattered from walls

You have two point sources of light of equal power + same height about ground
They are 10 feet apart

You stand a thick rod vertically in middle of two, and 4 feet in front of the point sources

What kind of shadow will be cast?

1) one shadow from one lamp?
2) two shadows from both lamps?
3) only intersection shadow from both lamps?
4) other?

2. Mar 30, 2012

### SHISHKABOB

depending on how bright the lights are, you'd see two shadows. If they were reaaaally bright, there'd be no shadows, but if they weren't so bright you'd see two shadows. Each shadow would be not as dim as would be if the other light weren't on.

I'm basing this on personal experience of remembering shadows on auditorium stages with multiple lights. Usually the object would cast as many shadows as there were lights above it, but the lights that were further away (or dimmer) would cast shadows that weren't as "strong" as the shadows cast by the lights that were closer (or brighter).

3. Mar 30, 2012

Almost....

With answer comes another question, but there are more than 2 shadows.

4. Mar 30, 2012

### DaveC426913

No. Whether Paul Muadid intended it or not, it's a trick question.

"...no light is scattered from walls."

:)

5. Mar 30, 2012

No light is scattered from walls, but the two point sources of light are generating light.

(in other words, the walls do not enter equation)

6. Mar 30, 2012

### DaveC426913

So, where do these shadows of which you speak get cast? :) But OK, let's grant the walls don't play in.

Each source casts a shadow of the rod and the other light source.

7. Mar 30, 2012

### SHISHKABOB

if no light is scattered from the walls, then how do you see the walls? And therefore how do you tell the difference between wall and shadow?

8. Mar 30, 2012

There is a floor, the shadow (whatever shadow that would be) would be cast on the floor.

9. Mar 30, 2012

### Staff: Mentor

You didn't say that there was a floor, or what color it was painted. All relevant information should be given in the problem statement, otherwise it's not a properly posed problem.

10. Mar 30, 2012

There's a ceiling, it too is black and non-reflective.

11. Mar 30, 2012

### DaveC426913

Perhaps the OP simply wishes to say "the walls reflect little enough light that any secondary reflections can be ignored".

12. Mar 31, 2012

A "shadow" counts as a separate shadow if it is from a different source OR if its gradient is different.
We're only interested in shadows cast onto floor.
There are three.

I'm still not sure why shadow 2 and 3 are not completely canceled out.
Shadow #2 is closer to left light source, so the left light source should completely cancel out the #2 shadow.
The same for #3.

Does it depend on how bright the light is?
Shadow #1 (the intersection of both shadows) is darker than #2 and #3

13. Mar 31, 2012

### DaveC426913

Where did you get this statement from? Source please.

Because shadows don't "cancel out". If you stood at a point in one of those shadows and took a light meter reading, how much light do you think you'd measure? Would it be less than you'd measure out in full light of both sources? Yes? Then you're in the shadow of at least one light regardless of how many other lights are shining on you..

One light source, one shadow. Two light sources, two shadows. The darker area is simply the intersection of the two shadows. It is not a third shadow.

BTW, the lights themselves will cast shadows of each other. Do you choose to ignore this?

14. Mar 31, 2012

### sophiecentaur

The words Umbra and Penumbra come to mind.
The problem isn't presented in an unambiguous way so it's hard to be sure what they're after.
I don't know how any shadow can "cancel out" another. The sum of the illumination 'outside' any shadow must be greater than the illumination within a shadow.