Optics - Having trouble with a few concepts

In summary: INSIDE the bowl of water, and the real image is the object's location outside the bowl of water?Yes, that is correct.
  • #1
Xyius
508
4
I am having trouble conceptualizing and drawing situations that involve glass, or any other medium that light travels through and doesn't totally reflect like mirrors. For example, say I have an object in a glass bowl of some sort. I know that the light rays will travel until they hit the glass. This is where I get confused. I know some of the light will travel through the glass bowl and some will reflect. I know that the ones that reflect will reflect with the same incident angle, except on the opposite direction of the normal. I also know that when the light passes through the glass and into the surrounding medium, it will refract assuming their refractive indexes are different. (Excluding the refractive properties in the glass itself.) AND I know that all the refracted beams will converge to a point. What I DON'T know is, what constitutes the image? By that I mean, if all the light rays converge outside the bowl, does that mean the image is seen outside the bowl? That doesn't make sense so it can't mean that.

As another example, if the bowl if full of water, the images tend to look bigger than they actually are. Is this because the light bounces off the sides and converges INSIDE the bowl some where closer to the wall? Why doesn't my drawing reflect that? If anyone could help me understand this I would appreciate it very much! :confused:

EDIT: Does the "Mirror Equation" apply here? Even though it isn't a mirror?
 
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  • #3
Xyius said:
AND I know that all the refracted beams will converge to a point...

You were doing o.k. up to here - this statement makes no sense. Also, there's this:

Xyius said:
As another example, if the bowl if full of water, the images tend to look bigger than they actually are. Is this because the light bounces off the sides and converges INSIDE the bowl some where closer to the wall?

This also does not make sense. Images of what? Real images, or virtual images that your cornea focuses onto your retina? It would help if you define what you are talking about - are you talking about viewing an object inside a bowl full of water with your eye?
 
  • #4
JeffKoch said:
are you talking about viewing an object inside a bowl full of water with your eye?

Yes. I guess it would be a virtual image. Although I am still unclear as to what is a real and virtual image.
EDIT:
It is my understanding that a virtual image would be in something like a mirror where the image looks as it if it behind the glass. And a real image is an image formed directly where on the point where the light rays converge. (As opposed to a mirror where they "converge" behind the mirror)
 
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  • #5
Xyius said:
Yes. I guess it would be a virtual image. Although I am still unclear as to what is a real and virtual image.
EDIT:
It is my understanding that a virtual image would be in something like a mirror where the image looks as it if it behind the glass. And a real image is an image formed directly where on the point where the light rays converge. (As opposed to a mirror where they "converge" behind the mirror)

If you draw a picture, a virtual image point is where you can backtrace every ray to a point in space that never was on all those rays. You can think of each ray as a line with a direction of travel in time - with a virtual image point, there never was a time when all those rays passed through that point, it just appears as if they did. So, an object placed in a bowl of water reflects light that hits it (however it got there) - those reflected rays are refracted at the water/bowl interface and again at the bowl/air interface, so these interfaces effectively form a lens that focuses the real object to a virtual image, and your eye can't tell the difference and thinks the virtual image position is where the actual object is. Depending on the distances, indices, etc., the virtual image can be magnified or demagnified.
 
  • #6
JeffKoch said:
so these interfaces effectively form a lens that focuses the real object to a virtual image

I think this is my problem, my ray trace diagrams aren't showing any possibility of light converging, be it real or virtual. I will post images of my drawings shortly.
EDIT: Actually I think I might be getting it! Just had a realization ha!

So I would indeed use the mirror equation to find the appeared position?
 
  • #7
Reduce the real object to a point, assume the bowl is perfectly spherical, and look at the problem in just two dimensions on a flat piece of paper. You need Snell's law and the refractive indices of the glass and the water - glass is typically 1.5, water is typically 1.3. Forget the reflected rays, they aren't important.
 

FAQ: Optics - Having trouble with a few concepts

1. What is the difference between reflection and refraction?

Reflection is the bouncing back of light rays from a smooth surface, while refraction is the bending of light rays as they pass through different mediums, such as air and water.

2. How does light interact with objects?

Light can be absorbed, transmitted, or reflected by objects. The color of an object is determined by the wavelengths of light that it reflects.

3. What is the relationship between wavelength and frequency?

Wavelength and frequency are inversely proportional. This means that as the wavelength increases, the frequency decreases, and vice versa.

4. What is the difference between convex and concave lenses?

A convex lens is thicker in the middle and thinner at the edges, causing light rays to converge and form a real image. A concave lens is thinner in the middle and thicker at the edges, causing light rays to diverge and form a virtual image.

5. How does the human eye see objects?

The cornea and lens of the eye refract light rays, which are then focused on the retina. The retina contains photoreceptors that convert light into electrical signals, which are then sent to the brain for interpretation.

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