# How long would it take for light to go through a human eyeball and back to the other side

I couldn't find any TAGS that fit my question... I don't know if I'm doing this correctly.

I want to know how long a "twinkling of an eye" would be. A twinkling of an eye is the time it takes from the moment the light hits the front of the eye, until it hits the back of the eye and is reflected back.

I googled some stuff and found that an average human eye is about 26.6666667 mm, so we'd double that to 53.3333334 mm, this is the distance light must travel.

the speed of light is 299,792,458 M per second, which is 299,792,458,000 mm per second. (notice i took the speed of light from Meters to millimeters).

So, now do i divide 299,792,458,000 mm by 26.6666667 mm? if so, google says that's: (299 792 458 000 millimeters) / (53.3333334 millimeters) = 5.62110858 × 109

so how do I turn 5.62110858 × 109 into time, or... i'm confused.. The smallest amount of time is a plank, or, $10^{-43}$ seconds. How close would the 'twinkling of an eye' be to a plank?

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This question was flagged for migration to physics.SE. However, as the FAQ states, math.SE will accept questions that are about "mathematical models and techniques" in physics, which I think this falls under. The OP is not asking about, for example, the physical basis for the reflection of light, which would be a question for physics.SE. – Zev Chonoles Jun 30 '11 at 12:32
A useful rule of thumb is that light travels roughly a foot per nanosecond. – David Speyer Jun 30 '11 at 13:42
This question was just simply about the math, and how to get the mathematical answer. – android.nick Jun 30 '11 at 14:39

I think I would define a "twinkling of an eye" to be the amount of time it takes for reflected light to travel from the surface of the eyeball in question to an observer, which would of course radically depend on how far away the observer was.

But, regarding your computation, your issue is just keeping track of units, otherwise known as dimensional analysis. $299,792,458,000\frac{\text{mm}}{\text{sec}}$ is not the same thing as $299,792,458,000\text{ mm}$.

You have a length, $L=26.\bar{6}\text{ mm}$, and a velocity, $V=299,792,458,000\frac{\text{mm}}{\text{sec}}$. To get a time quantity, you would have to divide $\frac{L}{V}$, because this corresponds to computing $$\frac{\text{length}}{\tfrac{\text{length}}{\text{time}}}=\frac{\text{length}\cdot\text{time}}{\text{length}}=\text{time}$$ It will have units of $$\dfrac{\text{mm}}{\tfrac{\text{mm}}{\text{sec}}}=\frac{\text{mm}\cdot\text{sec}}{\text{mm}}=\text{sec}.$$ So, the result is $$\frac{L}{V}\approx8.895\times 10^{-11}\text{sec}.$$

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So there are a couple of interesting things here. Firstly, a Plank time is about $10^{-43}$ seconds, not $10^{43}$. This is very, very small.

Ignoring all refraction and angles, when you say light goes about $3 * 10^{11} \frac{mm}{sec}$ and that you want to go about $3 * 10^1 mm$, you do the following: $$\frac{3*10^1 \; mm}{1} * \frac{1 \;sec}{3 * 10^{11}\; mm} = 10^{-10} \; sec$$

How close is this to a Plank time? Not at all. It covers about $10^{33}$ plank times. I'm having trouble finding a good metaphor for how big that is, but it's very large. When I find it, I'll update this answer.

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The human eyeball is not a vacuum. On the not unreasonable assumption that its average refractive index is much like that of water, the speed of light in the eye would decrease by a factor of about $1.33$ from its speed in a vacuum.

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