At a distance of 1 cm from the source, the flux from the isotropic source is one hundred times brighter. Which source is brighter fifty centimeters away? At what distance from the source are the two equally bright?
Could you explain the theory behind the answers too?
Isn't the flux for a collimated source independent of distance?
Well, well, well, looks like we've got ourselves a physics puzzle! I'm here to clown around and make it fun, so let's dive in!
First off, we've got an isotropic source, which means it emits light equally in all directions. And according to the problem, at a distance of 1 cm from the source, the flux is one hundred times brighter. That's quite a glow-up!
Now, let's think about this. Flux is a measure of the amount of light energy passing through a given area per unit time. So, when the flux is 100 times brighter at 1 cm, it means that at that distance, the same amount of light energy is spread over a much smaller area compared to when it's 50 cm away.
So, to answer the first part of your question, the source that's closer (at 1 cm) will be brighter. Let's pause for a round of applause for the little source that could!
But hey, don't fret! We still have the second part of the question. At what distance are the two equally bright? Well, since flux is dependent on the area, the flux will decrease with distance. And the good news is that we can determine the relationship between the decrease in flux and the increase in distance with a handy formula known as the inverse square law.
According to the inverse square law, the flux decreases with the square of the distance. In simpler terms, if you double the distance, the flux is only a quarter as bright. So, let's put on our math hats and figure out when the two sources will have equal brightness.
Since 1 cm is 50 times smaller than 50 cm, we need to square that ratio (50 times 50) to find out how much less bright the source will be at 50 cm. That's a whopping 2500 times less bright! So, the two sources will be equally bright when the one at 1 cm is 2500 times brighter than the one at 50 cm.
Now for the grand finale, we need to find the distance where the two sources are equally bright. Drumroll, please! By reversing the math, we can take the square root of 2500 to find that the answer is approximately 50.
So, to summarize: the source at 1 cm is brighter than the one at 50 cm, and the two sources will be equally bright at a distance of approximately 50 cm.
And as for your last question, whether the flux for a collimated source is independent of distance, the answer is yes! A collimated source emits light with parallel rays that do not diverge or converge. Therefore, the flux remains the same no matter the distance. No clowning around with this one!
I hope I've shed some light (pun intended) on your physics curiosity. Remember, laughter is the best way to brighten up any topic!
To compare the brightness of two sources at different distances, we need to understand the concept of flux and how it varies with distance. Flux, or luminous flux, is defined as the total amount of light emitted by a source per unit time.
The flux from an isotropic source (a source that emits light equally in all directions) decreases with distance according to the inverse square law. This means that as you move farther away from the source, the flux decreases proportionally to the square of the distance.
Now, let's analyze the given scenario:
1. At a distance of 1 cm from the source, the flux is one hundred times brighter:
This means that the flux at 1 cm is 100 times larger than the flux at some reference distance, let's call it "D_ref."
2. We are asked which source is brighter at a distance of 50 cm away:
To compare the brightness of the two sources, we need to determine the flux at this distance for both sources.
3. For an isotropic source, the flux decreases with the square of the distance:
So, the flux at 50 cm would be 1/2500 (since 50 cm = 2500 cm^2) of the flux at D_ref.
4. The flux at 1 cm is one hundred times brighter than at D_ref:
We can express this as a ratio: Flux(D_ref) = Flux(1 cm) / 100.
5. Using the inverse square law, we can set up the following equation:
Flux(50 cm) = Flux(D_ref) / 2500 = Flux(1 cm) / (100 * 2500)
6. To determine which source is brighter, we compare the flux values:
If Flux(50 cm) from source A is greater than Flux(50 cm) from source B, then source A is brighter. Otherwise, source B is brighter.
7. To find the distance at which the two sources are equally bright:
We need to find the distance, let's call it "D_equal," where Flux(D_equal) for both sources is equal.
8. Set up the equation:
Flux(D_equal)_A = Flux(1 cm)_B / 100 = Flux(D_equal)_B
9. Solve for D_equal:
Using the inverse square law, we can set up the following equation: Flux(D_equal)_A = Flux(D_equal)_B. Solve for D_equal.
In summary, the flux from an isotropic source decreases with the square of the distance. At 1 cm from the source, the flux is one hundred times brighter compared to some reference distance. Using the inverse square law, we can determine the flux at different distances. By comparing the flux at 50 cm for each source, we can determine which source is brighter. To find the distance where the sources are equally bright, we set the flux at that distance for each source equal to each other and solve for the distance.
To determine which source is brighter at a given distance and when the two sources are equally bright, we need to understand the concept of flux and how it relates to distance.
Flux is a measure of the flow of some quantity (such as light) passing through a given area. In the context of this question, flux refers to the amount of light emitted by the source that reaches a particular surface.
Regarding your question about collimated sources, it is correct that a collimated source produces a beam of light that stays parallel and does not diverge or converge. Due to this property, the flux from a collimated source remains constant regardless of the distance. However, in this question, the sources are described as isotropic, not collimated.
An isotropic source emits light uniformly in all directions, and its flux at a specific distance decreases with the square of the distance from the source. This behavior is known as the inverse square law.
Now, let's break down the problem step by step:
1. At a distance of 1 cm from the isotropic source, the flux is one hundred times brighter than at some reference distance. Let's call this reference distance D1.
Using the inverse square law, we can set up the equation: Flux (at 1 cm) = Flux (at D1) / (1 cm)^2.
2. Fifty centimeters away from the source, we need to determine which source is brighter. Let's refer to the first source (which has its flux at 1 cm as described above) as Source A, and the second source (unknown) as Source B.
Using the inverse square law, we can set up the equation: Flux (at 50 cm) from Source A = Flux (at D1) / (50 cm)^2.
We can also write the equation for Source B's flux at 50 cm: Flux (at 50 cm) from Source B = Flux (at D2) / (50 cm)^2.
3. To find out at what distance the two sources are equally bright, we can set up the equation: Flux (at D1) / (D1)^2 = Flux (at D2) / (D2)^2.
Solving this equation will give us the distance D2.
In summary, to find out which source is brighter at a certain distance and determine at what distance the two sources are equally bright, we need to use the inverse square law and set up the appropriate equations based on the given information.