# Thread: Question in physics for my homework

1. Dears:

I'm a dentist studying MSc - Dental Laser. we have a homework in the physics subject which I could not get the answer for. you may help pls

Q1: At which angle does a submerged diver in the ocean see the sunset? n Water = 1.33 (Hint: Sunset is when sun touches the horizon)

I applied Snell's Law and got the answer 48.6 considering that n Air = 1

but ... What I understood from this question is that the angel a Air is 90 degrees so why not an internal refraction occurs; I tend to think that when sun at the horizon point the diver should not be able to see the sun.

please try to explain this to me; you know we dentists are bad physics students!!

alaa

2. First, a reminder: when you give the answer, include "degrees" or "radians" with your angle depending on which you are using. So here your answer should be theta = 48.6 degrees.

Now, your question. Above the surface of the water, sunset occurs when the center of the sun reaches 90 degrees. At that point the light coming from the center of the sun is blocked by the surface of the earth, so there is no light ray coming towards you at exactly 90 degrees. So even though the light wouldn't be refracted into the ocean when it was traveling parallel to the surface, there is no ray traveling at that angle to consider anyway. In other words, as time goes on you see rays traveling up to a limit of 90 degrees, but not at that limit. So under the water, you will see light rays traveling over time up to a limit of 48.6 degrees, but not at that angle. The appearance is the same above and below the water except for the angle at which the sun disappears.

Good question. I had to think about it for a couple of minutes to get it visualized, which was fun. This problem is a good reminder that in physics it is always a good idea to keep in your mind a picture of what is going on.

3. thank you for the above reply.

If I got what you mentioned clearly, the light when theta air is 90 degrees, we will get the angle you mentioned.

but as zI know, and correct me if i'm wrong, there is no critical angle and total reflection when moving from lower density to a higher density medium (in this case air to water) so there should be rays refracting at the angle mentioned and this will be the angle at which the diver will see the sunset.
adding to the above, the actual position of the sun is different than the position we can see in the horizon during sunset; actually the virtual image of the sun is higher than the real position; so if the question is considering the virtual position when the sunset is happening then there will be a refraction and the diver will see the sun.

he will be able to see it till the virtual image of the sun disappears from the horizon, which is the position where the light will be blocked by the Earth.

initially when I thought about it and tried to draw it if thought that the diver will not be able to see the sunset unless the angle is very close to 90 but not exactly 90.

your comments please.

thanks;

4. Originally Posted by drsultan77
thank you for the above reply.

If I got what you mentioned clearly, the light when theta air is 90 degrees, we will get the angle you mentioned.

but as zI know, and correct me if i'm wrong, there is no critical angle and total reflection when moving from lower density to a higher density medium (in this case air to water) so there should be rays refracting at the angle mentioned and this will be the angle at which the diver will see the sunset.
adding to the above, the actual position of the sun is different than the position we can see in the horizon during sunset; actually the virtual image of the sun is higher than the real position; so if the question is considering the virtual position when the sunset is happening then there will be a refraction and the diver will see the sun.

he will be able to see it till the virtual image of the sun disappears from the horizon, which is the position where the light will be blocked by the Earth.

initially when I thought about it and tried to draw it if thought that the diver will not be able to see the sunset unless the angle is very close to 90 but not exactly 90.

your comments please.

thanks;
I think you have it. If the refraction of the atmosphere is also included, the sun appears to be a bit higher than its "real" position. Unless something was said explicitly in the problem about this, I doubt you were expected to take this into account.

I would say that technically you can't see the sun when it is at exactly the horizon line, but that is really an irrelevant issue. The refraction of light by the atmosphere, the visible size of the sun, and the irregular nature of the exact horizon, make issues of the visibility of the sun right at the horizon completely irrelevant in practice. But then, textbook problems are not always completely relevant.

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