Find Apparent depth due to a non-homogenous liquid

In summary, we have discussed the problem of finding the apparent depth of a vessel filled with a non-homogenous liquid with a varying refractive index. By using the paraxial approximation and relevant equations, we attempted to find the apparent depth by taking a differential strip at a certain height and using the formula for refractive index. However, integration was needed to solve the problem and further exploration is required. The conversation ended with the author stating they have found a solution.
  • #1
emailanmol
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THE ACTUAL PROBLEM:

A vessel of depth H is filled with a non-homogenous liquid whose refractive index varies with y as u=(2 -(y/H)), where y is measured from bottom of the vessel. Find the apparent depth as seen by an observer from above?
(Paraxial approximation is allowed)

RELEVANT EQUATIONS:

We know in paraxial approximation

u1/x=u2/y

Where u1 is refractive index of medium 1, x is object distance from surface, u2 is refractive index of surrounding medium , y2 is apparent depth.



MY ATTEMPT:

I took a differential strip of thickness dy at a height of y from bottom.
its refractive index is
(2-(y/H)-(dy/H)) and refractive index of element just below it is (2-(y/H))

Now,
Lets say that the image of the bottom of the vessel formed by refraction from all strips below this height y be at a distance x from the bottom.

Therefore, Its distance from the strip is y-x.

So using the formulae i mentioned.

(2-y/H)/(y-x)=
[(2-(y/H)-(dy/H))]/[y-x-dx)]

So ydx/H -2dx=-ydy/H+xdy/H.

After this am struck cause I cannot integrate.

I know other ways of solving this problem(mentioned in my textbook), but I want to know what exactly am I doing wrong here?
Any help/inputs will be really appreciated.
 
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  • #2
Never mind.I figured it out :-)

Admin can delete the thread
 

Related to Find Apparent depth due to a non-homogenous liquid

What is the concept of apparent depth due to a non-homogenous liquid?

The concept of apparent depth refers to how an object appears to be at a different depth when viewed through a non-homogenous or optically varying medium, such as a liquid. This is due to the bending of light as it passes through the medium, which causes the object to appear closer or further away than its actual position.

How is the apparent depth of an object in a non-homogenous liquid calculated?

The apparent depth of an object can be calculated using the formula: apparent depth = actual depth / refractive index of the medium. The refractive index is a measure of how much the light is bent as it passes through the medium, and it varies depending on the composition and density of the liquid.

What is the difference between apparent depth and actual depth?

Actual depth refers to the true physical depth of an object, while apparent depth is how deep the object appears to be when viewed through a non-homogenous medium. The difference between the two can be significant, especially when the refractive index of the medium is high.

How does the refractive index affect the apparent depth of an object in a non-homogenous liquid?

The refractive index of a liquid determines how much light is bent as it passes through the medium. A higher refractive index means that the light will be bent more, resulting in a greater change in the apparent depth of an object. This is why objects appear to be at a different depth when viewed through liquids with different refractive indices.

What are some examples of non-homogenous liquids that can cause a change in apparent depth?

Some common examples of non-homogenous liquids that can cause a change in apparent depth are water with different salt concentrations, alcohol with different sugar concentrations, and oil with different densities. Other examples include liquids with suspended particles or varying temperature gradients.

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