Why aren't the distances of the images the same in this problem?

In summary, the conversation discusses the correct method for finding the distance of an image to the surface of water in a scenario involving a spherical concave mirror and an object above the water line. The correct answer is 104cm, achieved by considering the refraction at the air/water interface and then the reflection in the mirror. The principle of reversibility is also mentioned, stating that if a source of light is placed at the location of the image, it will end up at the original location above the water line. The importance of considering the order of events in solving the problem is emphasized.
  • #1
jaumzaum
434
33
I was solving the below question

"An object is 30 cm above a container filled up with water. The lower end of the container is coated silver and acts like an ideal spherical concave mirror of radius 60cm. Find the distance of the image to the surface of water."

http://img856.imageshack.us/img856/5575/sadgsdfg.png [Broken]

If I find first find the image 1 made by the plan dioptre and than the image 2 that the image 1 makes with the mirror, I find 104cm (that is the correct answer)

But if I find first the image 3 that the mirror makes with the object (considering all in air) and THAN the image 4 that the image 3 does when going to water, I find 137cm.

Why the answers aren't the same? Shouldn't they be by the reversibility of light principle?
 
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  • #2
jaumzaum said:
If I find first find the image 1 made by the plan dioptre and than the image 2 that the image 1 makes with the mirror, I find 104cm (that is the correct answer)
I understand how you got the 104 cm answer. You considered the refraction at the air/water interface and then the reflection in the mirror.

But I don't understand the rest of what you did. The principle of reversibility tells you that if you now put a source of light at that spot, the image will end up at the original location above the water line. Which is true. Show your work so we can see what you did.
 
  • #3
Doc Al said:
I understand how you got the 104 cm answer. You considered the refraction at the air/water interface and then the reflection in the mirror.

But I don't understand the rest of what you did. The principle of reversibility tells you that if you now put a source of light at that spot, the image will end up at the original location above the water line. Which is true. Show your work so we can see what you did.

Hi Doc,

If I consider first the reflection:

1/30 = 1/170 + 1/s'

s' = 255/7cm => 140-255/7 = 725/7 cm under the surface

Now the refraction in the air /water surface

s'' = -725/7.4/3 = -138cm (138 cm under the surface)

Why the results are not the same?
 
  • #4
jaumzaum said:
Hi Doc,

If I consider first the reflection:

1/30 = 1/170 + 1/s'
Where did the 170 come from? The 'source' is now 104 cm under the water surface.
 
  • #5
Doc Al said:
Where did the 170 come from? The 'source' is now 104 cm under the water surface.

Thanks Doc, now that I thought better the way I'm following is not explained for the pronciple of reversibility :)

But anyway, I wanted to consider first the reflection and than the refraction (the 170 comes from the original distrance of the object to the mirror). I don't know why but for me the results should be the same. I thought that the order the things happenned were not important for the final result.So in a problem like this we should first consider the things that actually happened in first place and than the things that happenned after?
 
  • #6
jaumzaum said:
Thanks Doc, now that I thought better the way I'm following is not explained for the pronciple of reversibility :)
OK. I suspected I didn't understand you.
But anyway, I wanted to consider first the reflection and than the refraction (the 170 comes from the original distrance of the object to the mirror). I don't know why but for me the results should be the same. I thought that the order the things happenned were not important for the final result.So in a problem like this we should first consider the things that actually happened in first place and than the things that happenned after?
I would treat things in the order that the light meets them. Here the light first meets the air/water interface, them meets the mirror. (I see no reason why the answer would make sense if you mixed up the order of things.)
 
  • #7
Doc Al said:
OK. I suspected I didn't understand you.

I would treat things in the order that the light meets them. Here the light first meets the air/water interface, them meets the mirror. (I see no reason why the answer would make sense if you mixed up the order of things.)

Thanks Doc, now I've got it
I thought that I could mix up the order and get the same answer (I don't know why I thought it too) :)

[]'s
John
 

1. What is the reversibility of light?

The reversibility of light refers to the phenomenon of light being able to travel in both directions along the same path. This means that if light travels from point A to point B, it can also travel from point B back to point A.

2. How is light able to travel in both directions?

Light is an electromagnetic wave, which means it is made up of oscillating electric and magnetic fields. These fields can propagate in both directions, allowing light to travel back and forth along the same path.

3. Does this mean light can travel infinitely back and forth?

In theory, yes, light can continue to bounce back and forth between two points indefinitely. However, in reality, light can be absorbed or scattered by materials, which can limit its ability to travel back and forth.

4. Can the reversibility of light be observed in everyday life?

Yes, the reversibility of light can be observed in many everyday situations. For example, when you look in a mirror, light from your reflection is traveling back towards you, demonstrating the reversibility of light.

5. Are there any practical applications of the reversibility of light?

Yes, the reversibility of light is crucial in many technologies, such as fiber-optic communication, where light is used to transmit information back and forth along a fiber-optic cable. It is also used in various imaging techniques, such as mirrors and lenses, which rely on the reversibility of light to reflect and refract light in specific directions.

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