experimental-physicsglassopticsrefraction
There is a famous disappearing glass experiment in which you use pyrex glass (IoR=1.47) and oils with similar IoR to create an illusion that the glass disappeared.
Is there a liquid (mixture) that can reproduce this experiment with normal glass (IoR about 1.57)?
False.
To quote from the specific section in the list of common misconceptions:
Glass does not flow at room temperature as a high-viscosity liquid. Although glass shares some molecular properties found in liquids, glass at room temperature is an "amorphous solid" that only begins to flow above the glass transition temperature, though the exact nature of the glass transition is not considered settled among theorists and scientists. Panes of stained glass windows are often thicker at the bottom than at the top, and this has been cited as an example of the slow flow of glass over centuries. However, this unevenness is due to the window manufacturing processes used at the time. Normally the thick end of glass would be installed at the bottom of the frame, but it is also common to find old windows where the thicker end has been installed to the sides or the top. No such distortion is observed in other glass objects, such as sculptures or optical instruments, that are of similar or even greater age. One researcher estimated in 1998 that for glass to actually "flow" at room temperatures would take many times the age of the earth.
M2, is the paper and M1 is the coin. We are exerting a rightward force $F$ on the paper. It's obvious that the paper moves because of the force we are exerting on it, and it's acceleration is decreased by the friction force between the coin and the paper.(The paper is going rightwards, but the coin is exerting a friction force to the paper, leftwards). Based on Newton's third, because the coin is exerting a force on the paper leftwards, the paper is exerting a rightward force on the coin too. That's the reason the coin will stick to the paper when you're pulling the paper slowly. But the question is what happens when we increase this force. Firstly, the static friction force can not exceed a certain value. Which means if $F$ gets too high, the friction force can not keep up with $F$ and at some point the acceleration of the paper becomes more than the coin. So the coin can not keep up with the paper. So after a while, the coin just slips and falls into the glass.
I want to calculate the minimum force $F$ so that it causes the coin to slip. Imagine the maximum static friction force between the paper and the coin is $f_{k.max}$. So the maximum force that can be exerted on the coin is $f_{k.max}$. Which means the maximum acceleration of the coin is:
$a = F/M1 = \dfrac{f_{k.max}}{M1}$
So if you want to make the coin slip on the paper, you have to make the system move with an acceleration more that $a$. Which means:
$F > a * (M1 + M2) => F > \dfrac{(M1 + M2)f_{k.max}}{M1}$
If the force is below this value, then the coin doesn't slip, because the force exerted on it is not above $f_{k.max}$ but if the force exceeds this value, then $M1$ can not keep up with M2.
Think of the coin and the paper, as a whole system, only when the acceleration exceeds some value, the system breaks.
Forces on the paper:
$\sum{F_{paper}} = F - f_{k} = M_2 * a$
Forces on the coin:
$\sum{F_{coin}} = f_{k} = M_1 * a$
And we know that:
$f_k \leq f_{k.max} => M_1 * a \leq f_{k.max} => a \leq \dfrac{f_{k.max}}{M_1}$
$=> M_2 * a + f_{k} = F, => F \leq M_2 * \dfrac{f_{k.max}}{M_1} + f_{k.max} => F\leq \dfrac{(M_1 + M_2)f_{k.max}}{M_1} $
Best Answer
If the glass rod is surrounded by a liquid that has the same index of refraction, the speed of light will not change as it enters the rod meaning there'll be no refraction. If there is no refraction, the rod will appear "invisible". So the idea here is to get a liquid substance that has the same refractive index of glass, $n_g\approx 1.57$.
The table in this article lists, at standard temperature and pressure or STP (20$^\circ$ Celcius, 1 atmosphere pressure), the refractive index of benzene (C$_6$H$_6$) at $n_1 \approx 1.50$ and for carbon disulfide (CS$_2$) at $n_2\approx 1.63$ where$^1$ a mixture of these two liquids (or perhaps a mixture of any of the liquids in the correct proportions) should do the trick.
But note that the refractive index of a mixture of liquids is not obtained by the simple adding together of proportional volumes (to get the required refractive index), but rather $$n_{\text mix}^{\text id} =\frac{\phi_1\phi_2(n_1 -n_2)^2}{\phi_1n_1 +\phi_2 n_2 +\sqrt{\phi_1 n_1^2 +\phi_2 n_2^2}}$$ where $n_{\text mix}^{\text id}$ is the ideal refractive index for the mixture, and $\phi_1 , \phi_2$ are mixing fractions.
$^1$ It appears as though such a mixture, or any other mixture of liquids in the list, will be non-reactive at STP, but I cannot be too sure. C$_6$H$_6$ and CS$_2$ are both clear liquids that should inertly mix (but the prior is highly flammable and the latter is a neurotoxin and highly volatile!) at STP without leading to a liquid with noticeably different optical properties.