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A postage stamp is placed on a surface and a glass cube of refractive index $1.5$ is placed over it. When observed through the cube, the stamp appears at a height of $1.5cm$ from the bottom of the table. Another glass cube made of different material and same thickness is then placed over the first glass cube. This time when observed, the stamp appears at a height of $4cm$ from the bottom of the table. What is the refractive index of the second cube?

I started off by trying to figure out the height of the first cube with the following figure:

enter image description here

$EF$ is the postage stamp. I sampled two rays from the centre of the stamp at an angle of $60$ with the normal (ignore the $30$ in the figure).

Clearly, $\angle NIG = 30$.

Using Snell's Law, $n_{cube}\sin\angle NIG = n_{air}\sin\alpha$

$\implies \frac{3\sin30}{2} = \sin\alpha \implies \alpha = \sin^{-1}\left(\frac{3\sin30}{2}\right)$

Producing the emergent rays backward they intersect at a point $M$. It can be seen that $\tan\angle NMI = \frac{IN}{NM} \space (1)$

Let height of cube equal $h$. $MG = 1.5cm \implies NM = h - 1.5$

$\angle NMI = \alpha, \tan\angle NGI = \frac{IN}{d} \implies IN = d\tan60$

Putting all these together into $(1)$ we get,

$\tan\alpha = \frac{d\tan60}{d - 1.5} \implies d - 1.5 = \frac{d\tan 60}{\tan\alpha} \implies d - 1.5 = \frac{d\tan60}{\tan\left(\sin^{-1}\left(\frac{3\sin30}{2}\right)\right)}$

But, $\frac{\tan60}{\tan\left(\sin^{-1}\left(\frac{3\sin30}{2}\right)\right)} = 1.52752523165$, which would mean that $d - 1.5 = 1.5d$, which makes no sense.

I'll include the image with two blocks too, in case it aids in answering:

enter image description here

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  • $\begingroup$ Where did you read this problem? $\endgroup$ – user5402 Aug 18 '13 at 11:09
  • $\begingroup$ It's in my textbook. And the textbook is a fairly obscure one. You wouldn't have heard of it. Why? $\endgroup$ – Gerard Aug 18 '13 at 11:13
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I don't know if my solution is right, since it is much simpler!

The refractive index formula is

$$\mu = \frac{real \ depth}{apparent \ depth}$$

Now, let the thickness of the cube be $s$

For the bottom cube,

$1.5 = \frac{s}{s-1.5}$

Now, as seen by the top cube, the real depth is $s+s-1.5$. Note that the light doesn't know where it comes from(it just knows the direction), hence this is the real depth as seen by the top cube.

$\mu = \frac{2s-1.5}{2s-4}$

where $\mu$ is the required quantity.

Solving these two equations will give you your answer!

EDIT : Proof of the given formula :

image

From the image, it is clear that

$$\frac{real \ depth}{apparent \ depth} = \frac{\cot{r}}{\cot{i}} = \frac{\tan{i}}{\tan{r}}$$

Now, I am sorry to disappoint you with this bit. You have to make an assumption that the measurements are taken from right above the cube i.e. perpendicular to it's plane. This will give us that $i$ and $r$ are fairly small.

Hence,

$$\frac{real \ depth}{apparent \ depth} = \frac{\tan{i}}{\tan{r}} = \frac{\sin{i}}{\sin{r}} = \mu$$

is proved!

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  • $\begingroup$ That's all well and good. But where did the formula come from? Just using the formula blindly is no good is it? $\endgroup$ – Gerard Aug 20 '13 at 16:28
  • $\begingroup$ Yes, I know the derivation to the formula, and I assume that you know it too! But, once you know it, isn't it simply a waste of your time and brain power to use complex math. I am a man(teen actually) who goes by the advice of Feynman - Never let math limit your understanding of physics. $\endgroup$ – Cheeku Aug 20 '13 at 16:43
  • $\begingroup$ @Gerard Ok! Added! Sorry for a bad diagram! But that's all I could do in the limited time I got! $\endgroup$ – Cheeku Aug 21 '13 at 23:53
  • $\begingroup$ So, the formula @Cheeku posted makes an approximation for small angles. If you follow up from the $cot$ formula exactly, you end up in the formula I describe in my answer. By the way, I find the use of $cot$ very elegant. $\endgroup$ – chuse Aug 22 '13 at 14:05
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I believe you made a mistake at $\tan \angle NGI = \frac{IN}{d} \Rightarrow IN = d \tan 60$. It should be $\angle NGI = 30$ and not $\angle NGI = 60$. If you use $\angle NGI = 30$, you get $d - 1.5 = \frac{d \tan 30}{\tan(\arcsin\frac {3\sin 30}{2})}\approx 0.5 d $ and $d\approx 3.1 cm$.

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The problem of the poststamp is posing you the issue of lateral displacement, which occurs always that you have a light ray crossing parallel interfaces, from medium A to medium B and then back to medium A.

This may seem not to apply to your problem, since the poststamp is "touching" the glass; it doesn't make a difference, since you can think that there could be a little space of air between the glass and the poststamp (exact calculation will reveal the same). Also, your problem is stated in terms of vertical displacement; we'll see that later

So, what I propose is to calculate the lateral displacement induced by a pane of glass and then apply it to calculate directly the solution of your problem.

Lateral Displacement for Parallel Planes

enter image description here

Say you have a glass like the one in your figure (rectangular). Then, a ray will pass through it with a certain incidence angle, $\theta_1$. See that the angle with which the ray exits the glass has to be the same as the angle of incidence. Inside the glass, the ray travels with angle $\theta_2$, related with $\theta_1$ through Snell's law: $$ n_1\sin\theta_1=n_2\sin\theta_2$$ Analyze now the triangle that has as sides the displacement $d$, the real trajectory of the incident ray through the glass (the JG line in your drawing, but call it $l$ here), and the apparent trajectory. The angle between the real and the virtual trajectory will be $\theta_1-\theta_2$, so we can say that $$d = l \sin(\theta_1-\theta_2)$$ At this point, you have to replace the value of $l$, which is $$l = h \cos\theta_2$$ and you can expand the sine of the difference. You get: $$d = \frac{h}{\cos\theta_2} (\sin\theta_1\cos\theta_2-\cos\theta_1\sin\theta_2)$$ now use Snell for the sin and simplify the cosine: $$d = h \sin\theta_1 \left(1-\frac{n_1}{n_2}\frac{\cos\theta_1}{\cos\theta_2}\right)$$ You only have to replace now the $\cos\theta_2$ by $\sqrt{1-\sin^2\theta_2}$, and you will get $$d = h \sin\theta_1 \left(1-\frac{\cos\theta_1}{\frac{n_2^2}{n_1^2}-\sin^2\theta_1}\right)$$

Veritcal displacement Now that you got the lateral displacement, let's calculate the vertical one. Simply consider the triangle that the lateral and the vertical displacement ($d$ and $x$) forms, it's a right triangle where if you take angle $\theta_1$, the hypotenuse is $x$ and the opposite side is $d$, so $$x=d/\sin\theta_1$$ The whole formula is then $$x = h \left(1-\frac{\cos\theta_1}{\frac{n_2^2}{n_1^2}-\sin^2\theta_1}\right)$$

Your problem Use the vertical displacement of the first glass to get the value of $h$. Then use the vertical displacement of the second glass (4-1.5 cm) to get the value of $n_2$.

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