How Does Refraction Affect the Visibility of a Fish Inside a Spherical Bowl?

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Refraction significantly impacts the visibility of a fish inside a spherical bowl, with the apparent position and magnification calculated using the formula for lateral magnification. The focal length of the bowl is determined to be 7 cm, based on the relationship between radius and focal length, but there is confusion regarding whether this focal point lies within the bowl. Observers must consider the path of light rays and their convergence, which depends on the angle at which they enter the bowl. A proper ray diagram can clarify the relationship between the observer, the fish, and the focal point. Understanding these concepts is essential for accurately assessing the visibility of the fish in the bowl.
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Homework Statement



A small tropical fish is at the center of a water-filled, spherical fish bowl 28.0 cm in diameter.
(a) Find the apparent position and magnification of the fish to an observer outside the bowl. The effect of the thin walls of the bowl may be ignored. (b) Afriend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?

Homework Equations


The Attempt at a Solution



(sorry my poor english)

Since it is a plane interface, the lateral magnification is 1. Using m = (-na s' / nb s) and assuming that the bowl is one half of a sphere, so the object distante is 14 cm, I found for s' 10.52 cm. Then I divided 14 cm / 10.52 cm to find 1.33 (the book answer). But I'm not sure about that answer.

Second, I don't understand why the focal point isn't inside the bowl, if it's equal to 7 cm.
 
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Optics always makes more sense with a picture.
 
Plane interface?
May I ask how you got 10.52 cm
Why do you say the focal length is 7cm?
 
na (water) = 1.33
nb (air) ≅ 1
s = 14 cm (if the fish is at the bottom of the bowl and if it's one half of a sphere of radius 14 cm)
So s' = 10.52 cm.

The focal length is equal to R / 2 = D / 4 = 28 cm / 4 = 7 cm.
 
kent davidge said:
The focal length is equal to R / 2 = D / 4 = 28 cm / 4 = 7 cm.
How did you arrive at this? Which formula is this?
 
lol
Focal length is equal to radius / 2. It is what the theory says.
 
I think you should listen to Courtney and first draw a diagram.
Firstly do you still think it's a plane interface?
 
It would be like this?

30bflfa.jpg


but yet, I don't understand why the focal point is o
 
I am not convinced with the equation you've written at the bottom of that page
 
  • #10
It was an attempt to show that the focal point is at 7 cm. It doesn't make sense to say that the focal length is different of that value.
 
  • #11
kent davidge said:
lol
Focal length is equal to radius / 2. It is what the theory says.
That's for a mirror, so you can stop laughing, out loud or silently. How can the focal length of a lens not depend on the refractive index?
 
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  • #12
Kent, your formula of focal length being half of radius of curvature is valid for spherical reflecting surfaces,
This is refraction. Do you still feel that 7 is the focal length?
 
  • #13
kent davidge said:
It would be like this?

30bflfa.jpg


but yet, I don't understand why the focal point is o
That diagram makes no sense. You show P as at the centre of the sphere. An observer will always see the centre of the sphere as being on a straight line with the centre, not at some point P' off to the side.
 
  • #14
Oh thanks. I think because Engl is not my first language, I didn't know what is the actual shape of a bowl. I draw a new sketch where P and P' are in the same place and the focal point is outside the bowl.

szk85z.jpg
 
  • #15
kent davidge said:
Oh thanks. I think because Engl is not my first language, I didn't know what is the actual shape of a bowl. I draw a new sketch where P and P' are in the same place and the focal point is outside the bowl.

szk85z.jpg
You've drawn it as though P lies on the diameter that aligns with the observer, and the focal point is not on that. Is that what you intended?
 
  • #16
Yes. Is there any problem with this?
 
  • #17
kent davidge said:
Yes. Is there any problem with this?
Reread what I wrote in post #13. If you take a straight line from the observer through the centre of the sphere then the focal point for that observer must lie on that line. All points actually on the line will be seen as being on that line. Magnification consists of points off that line being seen as further from the line than they are.
 
  • #18
I thought focal point does not depend of the observer. So what really is this focal point? It's not just the local where the Sun's rays converge to?
 
  • #19
kent davidge said:
I thought focal point does not depend of the observer. So what really is this focal point? It's not just the local where the Sun's rays converge to?
A standard lens has its own axis. The focal point is taken as being on that axis because we base it on rays arriving parallel to that axis.
A sphere has no such natural axis. The focal point will depend on the direction your parallel rays come from.
In your diagram in post #14, point P appears to be the centre of the sphere, so the axis must pass through this point. You show rays from the sun arriving horizontally, so the focal point they converge to must be on the horizontal line through P.
Draw a horizontal line arriving above the axis. At entry to the sphere it will be refracted downwards, intersecting the axis at the focal point.
 
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  • #20
Does anybody know how to find the focal point? I am not sure how to solve this problem
 
  • #21
Meech__ said:
Does anybody know how to find the focal point? I am not sure how to solve this problem
Can you start by drawing a more accurate ray diagram than Kent managed?
 
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  • #22
I think there is an equation, I will try to solve it with this equation, and then I will come back
 
  • #23
Meech__ said:
Does anybody know how to find the focal point? I am not sure how to solve this problem
 

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  • #24
Here's the answer you'll need guys.
 

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