2023 A/L Physics Past Paper · Question 38 (English Medium)

Answer: (1) $\theta = \sin^{-1}\left(\dfrac{n_4}{n_1}\right)$

What the question asks

We want: the angle $\theta$ at the first interface, so that the ray goes along the interface of sheets 3 and 4.

What each letter means:

$\theta$: the angle of incidence at the interface of sheets 1 and 2, measured from the normal.

$\theta_2$, $\theta_3$: the angles the ray makes with the normal inside sheet 2 and inside sheet 3.

$n_1$, $n_2$, $n_3$, $n_4$: the refractive indices of the four sheets. They have no units.

Relevant theory

(2) For all rays of light of a given colour passing through two given media, the sine of the angle of incidence bears a constant ratio to the sine of the angle of refraction.

Hence, _bn_c = 1 / ( ₐn_b × _cnₐ ) = ₐn_c / ₐn_b

(3) On further increasing the angle of incidence, a situation is reached when the ray appears to pass along the interface separating the two media. The angle of incidence at this stage is called the "critical angle".

The first quote says that the sine of the angle of incidence divided by the sine of the angle of refraction is a constant, the refractive index. The second quote says how this index is found for two media from the indices of each. The third quote says that, at the critical angle, the ray passes along the interface.

For example, picture a line of marching soldiers who meet a patch of mud at an angle. The row that reaches the mud first slows down, so the whole line swings round.

Each new strip bends the line again, but the product of the index and the sine of the angle stays the same from strip to strip.

Required equations

$n_1 \sin\theta = n_2 \sin\theta_2$

$n_2 \sin\theta_2 = n_3 \sin\theta_3$

$n_3 \sin\theta_3 = n_4 \sin 90^\circ = n_4$

$\sin\theta = \dfrac{n_4}{n_1}$

Reasoning and calculation

The same number from sheet to sheet:

The sheets are parallel, so the angle at which the ray leaves one sheet is the angle at which it enters the next. At each interface the index of the new sheet over the index of the old sheet is $\dfrac{\sin i}{\sin r}$.

That gives $n_1 \sin\theta = n_2 \sin\theta_2$, then $n_2 \sin\theta_2 = n_3 \sin\theta_3$. The middle angles cancel, and $n_1 \sin\theta = n_3 \sin\theta_3$.

The ray along the last interface:

To go along the interface of sheets 3 and 4, the ray must be at the critical angle there. The ray leaves sheet 3 at $90^\circ$ to the normal, and $\sin 90^\circ = 1$.

So $n_3 \sin\theta_3 = n_4$. Put it in: $n_1 \sin\theta = n_4$, and so $\sin\theta = \dfrac{n_4}{n_1}$.

$0.75$

$\sin\theta = \dfrac{1.2}{1.6}$

$0.80$

$\sin\theta_2 = \dfrac{1.2}{1.5}$

$0.857$

$\sin\theta_3 = \dfrac{1.2}{1.4}$

$48.6$

$\theta = \sin^{-1}(0.75)$

$n4 / n1$

Reading the options

Now compare the form with the five options.

Option (1) is $\sin^{-1}\left(\dfrac{n_4}{n_1}\right)$. It matches.

With our numbers, option (2) gives $1.05$, option (3) gives $1.125$ and option (4) gives $1.575$. A sine cannot be bigger than $1$, so none of these three can match.

Option (5) gives $0.7$, not $0.75$. It does not match.

Checks

The refractive indices have no units, so the quantity inside $\sin^{-1}$ is a plain number, as it must be. It has to lie between $0$ and $1$.

For option (1) it is $\dfrac{n_4}{n_1} = 0.75$, which does.

Does it make sense?

If $n_4 = n_1$, then $\sin\theta = 1$ and $\theta = 90^\circ$. The ray would already be along the first interface. That is what we expect when the last sheet has the same index as the first.

The ray can run along the last interface when $n_3 > n_4$, because it goes from a denser sheet to a rarer one.

Conclusion

$\theta = \sin^{-1}\left(\dfrac{n_4}{n_1}\right)$

The ray goes along the interface of sheets 3 and 4 when $\theta = \sin^{-1}\left(\dfrac{n_4}{n_1}\right)$.

The answer is (1).

The quotes give the constant ratio of the sines, the index between two media from their indices, and the critical angle with the ray along the interface. The step $n_1 \sin\theta = n_2 \sin\theta_2$ at each interface follows from these quotes.

Teacher's note

Across parallel sheets, $n \sin(\text{angle})$ keeps the same value. The middle sheets drop out.

A ray along the last interface leaves at $90^\circ$, so $n_1 \sin\theta = n_4$.

Exam tip