EuPhO 2021, Теория — Задача 3. Glass ball

• b. 1.5

  Solution 1 gives n ≈ 1.498 ≈ 1.50; Solution 2 gives n_R ≈ 1.51; the grading scheme accepts answers within ±0.03.

• c. 0.013

  Solution 1 gives Δn ≈ 0.0132 ≈ 0.013; Solution 2 gives n_V − n_R ≈ 0.0145; the grading scheme accepts answers within ±0.003.

To begin with, let us notice that if a ray coming from a point $P$ on the stripe is refracted at point $Q$ at the surface of the ball towards a very distant point $A$ (which denotes the aperture of the camera lens), the ray will remain in the plane $PQO$ where $O$ is the centre of the sphere. This means that those rays which arrive from $P$ to $A$ must lay in the plane $POA$, and the rays can be conveniently depicted in the $POA$-plane.

The angle $\gamma = \gamma(\alpha)$ between vectors $PO$ and $QA$ is given by

and is a non-monotonous function of $\alpha$ which achieves its maximum $\gamma_0$ by a certain $\alpha_0$. This means that for a fixed $P$ and $A$, if $\angle OPA < \gamma_0$, there are two such angles $\alpha$ and, hence, two such positions $Q_1$ and $Q_2$ for the point $Q$ that the ray $PQA$ will reach the point $A$ (here we have assumed that the point $A$ is at a very big distance). So, when viewed from the point $A$, the image of the point $P$ is split into two points $Q_1$ and $Q_2$. On the other hand, if $\angle OPA > \gamma_0$, rays from $P$ cannot reach $A$, and such points on the stripe cannot be seen in the photo. If $\angle OPA = \gamma_0$, the two images $Q_1$ and $Q_2$ merge into a single point bridging the two images of a piece of stripe into a closed loop.

Now it becomes clear that the place where images $Q_1$ and $Q_2$ merge into a single image $Q_0$ is the key to finding the coefficient of refraction. Indeed, the point $Q_0$ can be found in the photo as the point where the blue ellipse (the images of a segment of the stripe in red light as the blue shadow is where the red light is missing) is touching a radius of the ball, see the figure below (we need to find tangent point with the radius because we need to consider plane $QOA$ which projects into a line through the ball’s centre). There is no difference between taking the tangent to the outer edge of the elliptical stripe and taking the tangent to the inner edge of it (point $Q_0'$ in the photo below). Then, $\sin\beta_0 = n\sin\alpha_0$ can be determined from the photo as the ratio of the lengths $h$ and $R$, $h/R \approx 0.765$, where $h$ denotes the distance of the ball’s centre $O$ from the line $Q_0A$, measurable in the photo as the length $OQ_0$ (when looking from the distant point $A$, we can see only the perpendicular-to-$OA$ component of the segment $OQ_0$, and $R$ is the ball’s radius.

Since we look for an extremum of $\gamma$, upon taking differentials from Eqns. (7,8), we obtain

from where

In order to find $\Delta n$, we could find in a similar way $n_V$ ($|OQ_0''| \approx 0.755R$, $n \approx 1.513 \approx 1.51$), but the result would have a huge relative uncertainty as $|OQ_0|$ and $|OQ_0''|$ have very similar lengths. A much more precise result will be obtained if we base our calculations on the segment length $|ST|$, see the figure below. $Q_0''$ is where the blue rays have an extremum for $\gamma$ while $Q_0'$ and $Q_0''$ are two red rays originating from the same point on the stripe. From the photo we can measure $|ST| \approx 0.20R$, hence we can use the small parameter $SQ_0''/R \approx 0.10$.

Rays from $S$, $T$, and $Q_0''$ arrive to the lens aperture, so all these rays (which originate from the same point $P$ on the stripe) have the same value of $\gamma$. So, we have a set of equations

and we would like to get an expression relating $\Delta n$ to $\sin\beta_{V1} - \sin\beta_{V2}$, where the indices 1 and 2 relate to the two different solutions. Expressing $\alpha_R = \alpha_V + \delta$ we obtain from Eq. (10) that $\beta_R = \beta_V + 2\delta$. Now, if we expand Eq. (12) into Taylor series, neglect the smallest term with $\Delta n\delta$, and keep in mind Eq. (11), we obtain

Here the two linear-in-$\delta$ terms cancel out due to Eq. (13) so that with $\sin\beta_V = n_V\sin\alpha_V$ we obtain

On the other hand, $|ST|/R = \sin\beta_{R1} - \sin\beta_{R2} \approx \sin(\beta_V + 2\delta) - \sin(\beta_V - 2\delta) = 4\delta\cos\beta_V$, hence

Grading scheme

A-i

• Proving that we can see a loop because for a range of points on the thread, a single point of thread creates two images: 0.7 pts
• The endpoints of that thread create one image forming thereby a closed loop: 0.3 pts

Else if only the fact that rays coming to A must be in the PQO plane is stated: 0.3 pts

A-ii Drawing a ray diagram where we can see that two rays originating from a single point arrive both to the lens while crossing the ball’s surface at different points and following the Snell’s law: 1.0 pt
Else if such idea is demonstrated by words or by a rough sketch (which does not obey Snell’s law): 0.5 pts

B-i Relating angle $\beta$ to the measurable distance $h$ in the photo (any point on the blue or red ellipse will earn the mark): 1pt

B-ii Measuring ratio of $h$ to $R$: 0.5 pts

B-iii The idea of using point $Q_0$: 1 pt (using $Q_0'$ will earn only 0.5 pts as this corresponds to the violet light)

B-iv Obtaining a correct expression of $\gamma$ as a function of $\alpha$ (or equivalent calculations in different parametrisation): 1 pt

B-v Finding extremum and hence, the final expression of $n$ as a function of $\beta$ or $h$: 1pt. If an incorrect expression is obtained from reasonable physics because of mistakes in algebraic manipulation: 0.5 pts

B-vi If the numerical answer is correct within $\pm 0.03$ and is found using reasonable physics: 0.5 pts. Else if the numerical answer is correct within $\pm 0.1$ and is found using reasonable physics: 0.3 pts. No point if the answer is guessed or is found using completely wrong or irrelevant physics.

C-i The idea of using the width or length of the blue ellipse: 1 pt (red ellipse cannot be used as we don’t know if the center of the red ellipse corresponds to the edge of the thread, or to a point at its middle)

C-ii Obtaining Eq (14) or something equivalent based on considering neighbouring red and violet rays: 1 pt. If an incorrect expression is obtained from reasonable physics because of mistakes in algebraic manipulation: 0.5 pts

C-iii Expressing $\Delta n$ correctly in terms of measurable quantities: 0.5 pts

C-iv Finding numerical answer correct within $\pm 0.003$: 0.5 pts. (0 pts if this answer is obtained without considering the width or height of the blue ellipse.)

Solution 2

One may alternatively prefer to consider a plane, perpendicular to the thread and containing the ball centre $O$ and the camera $A$. This is the plane of symmetry for the system and corresponds to the maximum width of the ovals. The same arguments as in the first Solution lead to

For $n < 2$ this function is not monotonous and may give the same result for two different values of $h$ (that is why the mapping of an arc is a closed oval).

To get the refraction index for red we solve the equation

which gives $n_R \approx 1.51$.

One could repeat the previous argument to find $n_V$. One could also compare the images of the same thread elements in different colours. This gives 4 equations

where $h_{1,4}$ means we can replace it with the value of $h_1$ or $h_4$. Similar rule applies for $h_{2,3}$. From each equation, we can solve for one value of $n_V$ and then take their average. This results in $n_V - n_R \approx 0.0145$.

Grading scheme

A-i

• Proving that we can see a loop because for a range of points on the thread, a single point of thread creates two images: 0.7 pts
• The endpoints of that thread create one image forming thereby a closed loop: 0.3 pts

Else if only the fact that rays coming to A must be in the PQO plane is stated: 0.3 pts

A-ii Drawing a ray diagram where we can see that two rays originating from a single point arrive both to the lens while crossing the ball’s surface at different points and following the Snell’s law: 1.0 pt
Else if such idea is demonstrated by words or by a rough sketch (which does not obey Snell’s law): 0.5 pts

B-i Relating angle $\beta$ to the measurable distance $h$ in the photo (any point on the blue or red ellipse will earn the mark): 1pt

B-ii Measuring ratio of $h$ to $R$: 0.5 pts

B-iii The idea of using points on the blue ellipse lying on the same radius of the sphere: 1 pt

B-iv Obtaining the correct equation to solve for $n_R$ numerically: 2 pts. If an incorrect equation is obtained from reasonable physics because of mistakes in algebraic manipulation: 1 pt

B-v If the numerical answer is correct within $\pm 0.03$ and is found using reasonable physics: 0.5 pts. Else if the numerical answer is correct within $\pm 0.1$ and is found using reasonable physics: 0.3 pts. No point if the answer is guessed or is found using completely wrong or irrelevant physics.

C-i The idea of using points on the red (or both) ellipse(s) lying on the same radius of the sphere: 1 pt

C-ii Obtaining at least one correct equation to solve for $n_V$ numerically: 1.5 pts. If an incorrect equation is obtained from reasonable physics because of mistakes in algebraic manipulation: 1 pt

C-iii Finding numerical answer correct within $\pm 0.003$: 0.5 pts.