Refraction: object disappearing in liquid

[subhradeep mahata]

Homework Statement

Why does an object disappear when it is placed in a liquid of refractive index equal to or greater than that of the object?

Homework Equations

The Attempt at a Solution

Well, I was just baffled by the problem. Finally, i came up with an answer: we see objects when they refract or reflect light. When refractive index of both are equal, there will be no refraction and hence the object won't be seen. However, i could not make out what happens to reflected light or what happens when refractive index of liquid is greater. Please help me out, if necessary you can use math.

 

[Charles Link]

The object in general will not disappear, but you can get something called total internal reflection, where because of Snell's law $n_1 \sin(\theta_1)=n_2 \sin(\theta_2)$, for incident angles/viewing angles $\theta_1$ such that $\frac{n_1}{n_2} \sin(\theta_1) >1$, you do get total reflection and thereby the object will appear as a mirror rather than being able to see through it and into it. Suggest you google "total internal reflection" for further info.

 

[subhradeep mahata]

In case of total internal reflection we are not able to see through the object?

 

[Charles Link]

In case of total internal reflection we are not able to see through the object?

The "internal" reflection in this case is really an external one. It's external to the object in the liquid. $\\$ On achieving a viewing angle that is such that you get complete reflection, you need to make sure the viewing angle rays aren't bent and thereby reduced below the critical angle by refraction: i.e. viewing a block (cube) of index $n_2$ inside a container of liquid of index $n_1$, you won't achieve this with a rectangular container (if the faces of the cube are parallel to to the faces of the rectangular container) when viewing from the air. Instead, if you have the container (transparent glass) be a cylinder/circular in shape, then the viewing rays (and likewise the observed rays) can pass straight through from your eye to the cube without any refractive bending. If the viewing angle exceeds the critical angle, the interface between $n_1$ and $n_2$ gets 100% reflection, and you have an ideal mirror. $\\$ For angles less than the critical angle, some of the light that you see will come from regions inside the cube and behind it. There will be a partial reflection at the interface, but it will not be 100% reflection. Basically, the fraction of transmitted energy plus the fraction of reflected energy add up to one: $T+R=1$. If $R$=100%=1 , then $T=0$.

 

[subhradeep mahata]

Ok , understood it...thanks

 

[Charles Link]

Please see the latest additions of $T+R=1$ in post 4.

 

[subhradeep mahata]

I saw it...so basically if the n of liquid is greater than n of object, we can't see it through air, as it behaves as an ideal plane mirror for i>= c ?

 

[Charles Link]

I saw it...so basically if the n of liquid is greater than n of object, we can't see it through air, as it behaves as an ideal plane mirror for i>= c ?

You need to make sure by doing a ray trace from your eye to the $n_1 - n_2$ interface, that $\theta_i>\theta_c$. If that is the case, the interface will be 100% reflective and you will not see what is behind it. $\\$ An example of this can be seen if you view a rectangular aquarium filled with water, and look through the front of the aquarium and view e.g. the left end face of the aquarium. You will find viewing angles for which you can not see through to whatever is behind the left end face. The left end face will become a perfect mirror past a certain viewing angle.

 

[subhradeep mahata]

Ok, my doubt is cleared. Thanks again.

 

[VKint]

Actually, I think the point is that total internal reflection can't occur in this situation.

Picture the object as a block of acrylic (for example) submerged in a fluid. If the object is visible from above the fluid, then there must be a path for a ray of light to travel via refraction through the fluid, strike the block at some angle $\theta$ (measured relative to the surface normal), and reflect upward back towards the surface. If $\theta$ is too small, then the light will (mostly) refract through the block rather than reflect, rendering the block (mostly) "invisible" at this angle.

So the first thing we want to do is calculate this "critical angle" below which light will tend to refract through the block. Let $n_1$ and $n_2$ denote the indices of refraction of the fluid and block, respectively. If $\alpha$ is the angle at which light is refracted through the block, Snell's Law gives $n_1 \sin(\theta) = n_2 \sin(\alpha)$, or $\sin(\alpha) = (n_1 / n_2) \sin(\theta)$--in particular, if $\theta > \theta_c = \sin^{-1}(n_2 / n_1)$, then no refraction is possible because Snell's Law has no solution for $\alpha$. (This, by the way, is the condition for total internal reflection that Charles Link alluded to.)

Next, we want to show that any ray of light refracted from above the fluid's surface will strike the block at an angle less than the critical angle $\theta_c$ (again, measured from the surface normal). Suppose we have a ray that strikes the block at angle $\phi > \theta_c$. Assuming the block's surface is parallel to the fluid's, simple geometry implies that the ray must have been refracted at the same angle $\phi$ through the surface of the fluid. This means that the angle of incidence $\phi_0$ of our hypothetical ray on the fluid must obey the following relation: $$\sin(\phi_0) = n_1 \sin(\phi) > n_1 \sin(\theta_c) = n_2$$ (I'm assuming the ambient environment to be air or vacuum, so that the index of refraction outside the fluid is roughly 1.) If $n_2 > 1$ (and it would be strange indeed for the block to have an index of refraction less than that of air), then this relation is impossible to satisfy.

To summarize: Any ray of light passing from the air above the fluid to the block submerged in the fluid will strike the block at an angle less than $\theta_c$, precluding the possibility of total reflection from the block. However, this is not quite the end of the story.

Even without total reflection, some light will still be reflected by the surface of a dielectric material--at least up to a second "critical angle" called Brewster's angle. Without going into details, Brewster's angle is given by $\theta_B = \tan^{-1}(n_{ref}/n_{inc})$ for light reflected from a medium with index $n_{ref}$ and traveling through a medium with index $n_{inc}$. Light incident on the reflecting medium will be totally refracted (i.e., with no reflection) if the angle of incidence is less than $\theta_B$. So once again, we could ask whether any ray of light can strike the block at an angle greater than Brewster's without violating Snell's Law.

Unfortunately, the answer seems to be yes--it is possible to have $\theta_B < \theta_c$, and moreover the equation $n_1 \sin(\theta_B) = \sin(\phi_0)$ does seem to have solutions. (I'll let you think about when and why this might be true.) In this case, the block should be visible for a small band of incident angles .

 

[Charles Link]

@VKint That's why I mentioned in post 4 in the second paragraph that ideally you need a circular container, or alternatively, tilt the cube at an angle inside a rectangular container.

 

[VKint]

Right, sorry--when I started writing my post, you'd only posted one reply. Too slow :(

 

[VKint]

Still, I think the simple case of a rectangular object submerged parallel to the surface of a fluid might be what the problem is getting at. In this case, most of the light that strikes the block will be refracted, leading to a sort of "invisibility" effect.

 

[Charles Link]

Still, I think the simple case of a rectangular object submerged parallel to the surface of a fluid might be what the problem is getting at. In this case, most of the light that strikes the block will be refracted, leading to a sort of "invisibility" effect.

For $n_1 \approx n_2$, regardless of whether $n_1>n_2$ or $n_1<n_2$, the block will appear to be very transparent. Yes, in that case it could also be described as nearly "invisible".

 

[Charles Link]

And a follow-on to post 14: It may be of interest to the OP that at normal incidence the energy reflection coefficient $R=\frac{(n_1-n_2)^2}{(n_1+n_2)^2}$, and that basically quantifies the result, regardless of whether $n_1 >n_2$ or $n_1<n_2$. And the energy transmission coefficient is $T=1-R$.

 

[subhradeep mahata]

Thanks to both Charles Link and VKint for helping me out with brilliant answers ! :)