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If I poured water into my tea, would I see more or less of the bottom of the tea-cup?

Intuitively, there would be as many particles blocking as many photons, and so I'd see the bottom just as clearly as before.

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If the water gets mixed with the tea and no chemical change occurs, you can't see the bottom more clearly. However, before it gets mixed up, you may create regions in the liquid that contain a smaller density of the impurities (tea), and that's where you can see further, while you cannot see as much through the denser, more "contaminated" columns. – Luboš Motl Feb 6 '11 at 20:01
@Ludos: I hadn't thought of the immediate effect of impurites :) – user1778 Feb 6 '11 at 22:07
up vote 4 down vote accepted

There is some misconception in Tims question already:

as many particles blocking as many photons, and so I'd see the bottom just as clearly as before

Blocking of photons will not impair clarity at all!

One has to distinguish absorption and scattering! Tea is a solution of some dyestuffs in water (optically) this dyestuff absorbs light of certain wavelengths (making the tea looking yellow/brown) but the tea is not made turbid ("Unclear") by that absorption. (Think of some color filter for cameras!)

The absorption is governed by Lambert/Beer law. The law shows, that doubling the length of the liquid column is "neutralized" by reducing concentration to half of original value.

Now the question of turbidity, In practice a tea may contain some particles from lime in the water, often some agglomerates from lime and some tanning agents in the tea. On top there is some fat/oil which drift as droplets or collect on the surface.

Any particle floating in the tea, which is about 50 nm up to some µmeters and having a index of refraction different enough from the index of water, will cause some scattering.

This scattering will blur the picture.

Most of the scattering in such suspensions like tea or milk are governed by Mie scattering. Scattering/concentrations/length is not as simple as Lambert-Beer, but at low concentrations one can assume an analog dependence.

Some practical note: when having a good (low lime) water and and the leaves are in a bag with good filter action, and You wait for a moment until the oil accumulates at the surface, tea will be a nearly perfectly clear liquid.


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Thank you for the extensive answer, Georg. It seems that I should distinguish better between clarity and turbidity. Do you suggest that the Mie scattering would be the main reason that I cannot see the bottom of a coffee cup? (If I may substitute the liquid in my original question as a result of your practical note.) Since you say that the Mie scattering is not linear: would it become clearer or not from watering it down? – user1778 Feb 6 '11 at 21:43
+1, but can you please follow standard formatting? – mbq Feb 6 '11 at 23:11
Hello Tim, I said, at low concentrations, scattering can be aproximated by Lambert-Beer. This would be the case in tea, whereas for some milk this would not work. Re. Coffee, no for a freshly brewn cup coffee is a solution of rather dark brown dyestuffs, but nevertheless clear. After some hours a coffee becomes turbid. You will see this easily when You thin down such an 2old" coffee. – Georg Feb 6 '11 at 23:11
Hello mbq, what is that "standard"? When I do not the extra lines, the text flows on, not having the lines I want. What can I do? – Georg Feb 6 '11 at 23:14

You are right -- there is an approximated formula for this, called Lamber-Beer law:

$A=\epsilon l c$,

where $A$ is absorbance (log of quotient of light intensities before and after the solution), $\epsilon$ is substance dependent coefficient (constant), $l$ is the size of layer of solution and $c$ is the concentration. Let's now say that the glass cross-section is $S$, and initial height of fluid inside is $h$; adding water up to height $\alpha h$ will decrease concentration $c$ $\alpha$ times, but increase $l$ $\alpha$ times, so absorbance will stay the same.

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Thank you, I was looking for something like Lambert-Beer. The Wikipedia article mentions that a condition for its validity is that "the absorbing medium must not scatter the radiation". When would that be the case? – user1778 Feb 6 '11 at 21:22
@Tim Of course not ;-) But in most real-world conditions the corrections are small and this formula works pretty well. – mbq Feb 6 '11 at 23:04

Index of refraction is related to density, as your intuition predicted. The question is how would tea and water look like at a molecular level. Without a microscope I can only guess, of course, but the tea as being big solid chunks floating around in the water might explain.

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Index of refraction would have very little to do with this. – Colin K Feb 6 '11 at 20:23

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