Minimum height of water to make the particle visible

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Homework Help Overview

The discussion revolves around determining the minimum height of water required for an observer to see a submerged particle. The subject area includes concepts of optics, specifically refraction and the application of Snell's law.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to use normal shift to solve the problem, raising questions about the validity of this approach given the geometry of the situation. Participants discuss the implications of viewing angles and the limitations of the normal shift formula.

Discussion Status

Participants are actively engaging with the original poster's reasoning, validating some concerns while suggesting alternative methods, such as deriving a more general equation for the shift. There is a recognition of the need to explore the relationship between light refraction and the observer's perspective.

Contextual Notes

There are constraints related to the physical setup, including the height of the container and the assumptions made about viewing angles. The original poster's calculations are questioned based on these constraints.

palaphys
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Homework Statement
A cylindrical vessel, whose diameter and height both are
equal to 30 cm, is placed on a horizontal surface and a
small particle P is placed in it at a distance of 5.0cm
from the centre. An eye is placed at a position such that
the edge of the bottom is just visible (see figure 18-E8).
The particle P is in the plane of drawing. Up to what
minimum height should water be poured in the vessel
to make the particle P visible ?
Relevant Equations
snells law
1759721321523.webp

I know how to solve this using Snell's law and geometry, but I thought of a different approach- using normal shift
Firstly here is a diagram for the geometry of the situation:
1759721583683.webp

Now somehow, if we raise the image of P to a height of ##h## from the bottom, it will be right on the line of sight of the observer, so technically he would be seeing that. We know that if we fill the beaker with water, the object will appear at a higher position.
so if we use the formula for normal shift, assuming the length of the water column to be ##x##,
## 10=x(1-3/4) ## (assuming refractive index of water to be 4/3)
## x= 40cm##
which is wrong, I think due to the following reasons
i) I remember that this formula was derived only for near normal viewing. but here it is not the case.
ii)the height of the container itself is only 30cm, then how would we fill it up to 40cm

My question is, are my reasons valid? are they conceptually sound?
 
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Your reasons for why it doesn't work are valid. Note that the observing eye as shown will still see the bottom of the container shifted up. There is a more general equation that you can derive for that shift that works for lines of sight appreciably away from the normal. See if you can derive it. Check your work by showing that it reduces to the "normal shift" equation when the angle away from the normal is small.
 
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Think about how light bends when it passes from water to air (refraction).
 
realJohn said:
Think about how light bends when it passes from water to air (refraction).
As stated in post #1, @palaphys knows how to solve the problem. The question being asked of the forum is why the suggested alternative method does not work.
An interesting corollary is that a flat base viewed through a uniform depth of water does not appear flat. So what shape does it appear to be?
 
Last edited:
haruspex said:
As stated in post #1, @palaphys knows how to solve the problem. The question being asked of the forum is why the suggested alternative method does not work.
Thank you.

Normal shift doesn't necessarily sound correct to me. This would only work when you're looking straight down. Just use both Snell's law and geometry.
 

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