Homework Help: A Spherical Fish bowl

1. Jan 24, 2016

kent davidge

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

3. 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 dont understand why the focal point isn't inside the bowl, if it's equal to 7 cm.

2. Jan 24, 2016

Dr. Courtney

Optics always makes more sense with a picture.

3. Jan 24, 2016

Suraj M

Plane interface?
May I ask how you got 10.52 cm
Why do you say the focal length is 7cm?

4. Jan 24, 2016

kent davidge

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.

5. Jan 24, 2016

Suraj M

How did you arrive at this? Which formula is this?

6. Jan 24, 2016

kent davidge

lol
Focal lenght is equal to radius / 2. It is what the theory says.

7. Jan 24, 2016

Suraj M

I think you should listen to Courtney and first draw a diagram.
Firstly do you still think it's a plane interface?

8. Jan 24, 2016

kent davidge

It would be like this?

but yet, I dont understand why the focal point is o

9. Jan 24, 2016

Suraj M

I am not convinced with the equation you've written at the bottom of that page

10. Jan 24, 2016

kent davidge

It was an attempt to show that the focal point is at 7 cm. It doesnt make sense to say that the focal lenght is different of that value.

11. Jan 24, 2016

haruspex

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?

12. Jan 24, 2016

Suraj M

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. Jan 24, 2016

haruspex

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. Jan 24, 2016

kent davidge

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.

15. Jan 24, 2016

haruspex

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. Jan 24, 2016

kent davidge

Yes. Is there any problem with this?

17. Jan 24, 2016

haruspex

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. Jan 25, 2016

kent davidge

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. Jan 25, 2016

haruspex

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.

20. Dec 1, 2016

Meech__

Does anybody know how to find the focal point? I am not sure how to solve this problem

21. Dec 1, 2016

haruspex

Can you start by drawing a more accurate ray diagram than Kent managed?

22. Dec 1, 2016

Meech__

I think there is an equation, I will try to solve it with this equation, and then I will come back

23. Jun 16, 2017

AleyX98

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24. Jun 16, 2017

AleyX98

Here's the answer you'll need guys.

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