Hong Kong Advanced Level Examination (HKALE)

August 31, 2023September 2, 2023

202308311431 Solution to 2007-AL-PHY-IIA-9

A diver at a depth of $d$ below the water surface looks up and finds that the sky appears to be within a circle of radius $r$ . Which of the following correctly gives the expression for the critical angle?

A. $\tan c=\displaystyle{\frac{r}{d}}$
B. $\sin c=\displaystyle{\frac{r}{d}}$
C. $\tan c=\displaystyle{\frac{d}{r}}$
D. $\sin c=\displaystyle{\frac{d}{r}}$

Background.

The critical angle is the smallest angle of incidence that yields total reflection, or equivalently the largest angle for which a refracted ray exists.

_{Wikipedia on Critical angle (optics)}

The light travelling from air to water cannot suffer total internal reflection because for total internal reflection, the essential condition is that light should travel from a denser medium to a rarer medium with incidence angle more than the critical angle.

_{Toppr on Total internal reflection}

Roughwork. (making a short story long)

Subscripts $\text{}_a$ and $\text{}_w$ stand for air and water respectively.

$\begin{aligned} n_w\sin \theta_w & = n_a\sin\theta_a \\ \textrm{Eq. (1):\qquad }\frac{n_w}{n_a} & = \frac{\sin\theta_a}{\sin\theta_w}} \\ \frac{\mathrm{d}}{\mathrm{d}t}\bigg( \frac{n_w}{n_a} \bigg) & = \frac{\mathrm{d}}{\mathrm{d}t}\bigg( \frac{\sin\theta_a}{\sin\theta_w}\bigg) \\ 0 & = \frac{\dot{\theta}_a\sin\theta_w\cos\theta_a-\dot{\theta}_w\sin\theta_a\cos\theta_w}{\sin^2\theta_w} \\ \textrm{Eq. (2):\qquad }\frac{\dot{\theta}_a}{\dot{\theta}_w} & = \frac{\tan\theta_a}{\tan\theta_w}=\textrm{non-constant}\\ \end{aligned}$

At some height $h$ above the water surface, where $\theta_a$ is the angle of refraction:

$\begin{aligned} \textrm{Eq. (3):\qquad }\tan\theta_a & = \frac{r_a}{h} \\ \textrm{where\qquad }\theta_a & \in [0,90^\circ] \\ \textrm{and \qquad }h & = \textrm{Const.}\in [0,\infty ) \\ r_a & = r_a(\theta_a) \in [0,\infty )\\ \end{aligned}$

To some depth $d$ beneath the water surface, where $\theta_w$ is the angle of incidence:

$\begin{aligned} \textrm{Eq. (4):\qquad }\tan\theta_w & = \frac{r_w}{d} \\ \textrm{where\qquad }\theta_w & \in [0,90^\circ] \\ \textrm{and \qquad }d & = \textrm{Const.}\in [0,\infty ) \\ r_w & = r_w(\theta_w ) \in [0,\infty ) \\ \end{aligned}$

Differentiating $\textrm{Eq. (3)}$ and $\textrm{Eq. (4)}$ :

$\begin{aligned} \textrm{Eq. (3)':\qquad }h\dot{\theta}_a\sec^2\theta_a & = \dot{r}_a \\ \dot{\theta}_a & = \frac{\dot{r}_a}{h\sec^2\theta_a} \\ \end{aligned}$

$\begin{aligned} \textrm{Eq. (4)':\qquad }d\dot{\theta}_w\sec^2\theta_w & = \dot{r}_w \\ \dot{\theta}_w & = \frac{\dot{r}_w}{d\sec^2\theta_w} \\ \end{aligned}$

Dividing $\textrm{Eq. (3)'}$ by $\textrm{Eq. (4)'}$ :

$\textrm{Eq. (5):\qquad }\displaystyle{\frac{\dot{\theta}_a}{\dot{\theta}_w} =\bigg(\frac{\dot{r}_a}{\dot{r}_w}\bigg) \bigg(\frac{d}{h}\bigg)\bigg(\frac{\sec^2\theta_w}{\sec^2\theta_a}\bigg)}$

Equating $\textrm{Eq. (2)}$ and $\textrm{Eq. (5)}$ :

$\begin{aligned} \textrm{Eq. (6):\qquad }\frac{\tan\theta_a}{\tan\theta_w} & =\bigg(\frac{\dot{r}_a}{\dot{r}_w}\bigg) \bigg(\frac{d}{h}\bigg)\bigg(\frac{\sec^2\theta_w}{\sec^2\theta_a}\bigg) \\ \frac{\tan\theta_a}{\tan\theta_w} & = \frac{(r_a/h)'}{(r_w/d)'}\bigg(\frac{\sec\theta_w}{\sec\theta_a}\bigg)^2\\ \frac{\tan\theta_a}{\tan\theta_w} & = \frac{(\tan\theta_a)'}{(\tan\theta_w)'}\bigg(\frac{\sec\theta_w}{\sec\theta_a}\bigg)^2\\ \frac{\tan\theta_a}{\tan\theta_w} & = \bigg(\frac{\dot{\theta}_a}{\dot{\theta}_w}\bigg)\bigg(\frac{\sec\theta_a}{\sec\theta_w}\bigg)^2\bigg(\frac{\sec\theta_w}{\sec\theta_a}\bigg)^2=\frac{\dot{\theta}_a}{\dot{\theta}_w}\\ \end{aligned}$

This $\textrm{Eq. (6)}$ is redundant as it is the same as $\textrm{Eq. (2)}$ . Never mind. But rewrite $\textrm{Eq. (6)}$ (/ $\textrm{Eq. (2)}$ ) as follows:

$\begin{aligned} \dot{\theta}_w\cot\theta_w & = \dot{\theta}_a\cot\theta_a \\ \frac{\mathrm{d}}{\mathrm{d}t}\big(\ln |\sin \theta_w|\big) & = \frac{\mathrm{d}}{\mathrm{d}t}\big(\ln |\sin \theta_a|\big) \\ \ln|\sin\theta_w| & = \ln|\sin\theta_a| \\ \theta_w & = k\pi -\theta_a \qquad (k\in\mathbb{Z})\\ \textrm{and}\qquad \theta_a & \geq \theta_w \\ \end{aligned}$

Note that refractive index for air is (approximately) $n_a=1$ ,

$\begin{aligned} \textrm{Eq. (7):\qquad }\frac{\sin\theta_a}{\sin\theta_w} & =n_w \\ \sin\theta_a & = n_w \sin\theta_w \\ \sin\theta_a & = n_w\sin (90^\circ-\theta_a) \\ \sin\theta_a & = n_w\cos\theta_a \\ \textrm{Eq. (7)':\qquad }n_w & =\tan\theta_a\\ \sin (90^\circ -\theta_w) & = n_w\sin\theta_w \\ \cos\theta_w & = n_w\sin\theta_w \\ \textrm{Eq. (7)'':\qquad }n_w & = \frac{1}{\tan\theta_w}} \\ \end{aligned}$

Equating $\textrm{Eq. (7)'}$ and $\textrm{Eq. (7)''}$ :

$\begin{aligned} \tan\theta_a & = \frac{1}{\tan\theta_w} \\ \tan\theta_a\tan\theta_w & = 1 \\ \textrm{Eq. (8):\qquad }\tan (90^\circ -\theta_w)\tan\theta_w & = 1 \\ \end{aligned}$

Taking limit on $\textrm{Eq. (8)}$ ,

$\begin{aligned} \Big(\lim_{\theta_w\to c}\tan (90^\circ -\theta_w)\Big)\Big(\lim_{\theta_w\to c}\tan\theta_w\Big) & =1 \\ (\cot c)\bigg(\frac{r}{d}\bigg) & = 1 \\ \tan c & =\frac{r}{d} \\ \end{aligned}$

And the answer is A.

January 11, 2023January 11, 2023

202301110923 Solution to 1980-AL-PHY-IA-26

A charged ball $X$ is suspended by a string.

When a uniform electric field $E$ is applied horizontally, the ball is displaced a horizontal distance $a$ such that the string makes an angle $\theta$ with the vertical.

Roughwork.

As always start with a free-body diagram. Hence, draw

Resolving components, write

$\begin{cases} \textrm{Horizontal:}\quad & T\sin\theta = qE \\ \textrm{Vertical:}\quad & T\cos\theta = mg \\ \end{cases}$

Squaring and summing, write

$\begin{aligned} (T\sin\theta )^2+ (T\cos\theta)^2 &= (qE)^2 + (mg)^2\\ T & = \sqrt{q^2E^2+m^2g^2} \\ \end{aligned}$

By considering the electric field strength $E$ as an independent variable, and the angle of elevation $\theta$ the dependent variable, make differentiation

$\displaystyle{\mathrm{d}E = \bigg(\frac{T\cos\theta}{q}\bigg)\,\mathrm{d}\theta}\qquad(\theta\in [0,90^\circ])$

This problem is not to be attempted.

February 8, 2022October 24, 2022

202202081107 Solution to 1980-AL-PHY-I-33

In an experiment, a student takes several readings of the variable $x$ and $y$ . He plots a graph of $\log y$ against $\log x$ and finds that it is a straight line of gradient $m$ , making an intercept $c$ on the positive $\log y$ axis.

What is the relation between $x$ and $y$ ?

Solution.

$\begin{aligned} \log y & = m\log x+c\\ \log y & = \log x^m + c\\ \log y-\log x^m & = c\\ \log \bigg( \frac{y}{x^m}\bigg) & = c \\ \cdots\enspace \textrm{let }c =\log a&\enspace \cdots\\ \log \bigg( \frac{y}{x^m}\bigg) & = \log a \\ \frac{y}{x^m} & = a\\ y & = ax^m \\ \end{aligned}$

September 16, 2019October 24, 2022

201909160458 Solution to 2001-AL-PHY-IIA-5

(bad example)

The distance between $X$ and $Z$ is $s=ut+\frac{1}{2}at^2$ .

The mid-point $Y$ of $XZ$ is $\frac{1}{2}s$ away from $X$ after some time $t'<t$ .

$\begin{aligned} \frac{1}{2}s & =ut'+\frac{1}{2}at'^2 \\ \frac{1}{2}\bigg( ut+\frac{1}{2}at^2 \bigg) & =ut'+\frac{1}{2}at'^2 \\ 0 & = \bigg( \frac{1}{2}a\bigg) t'^2 + (u)t' + \bigg( -\frac{1}{2}ut-\frac{1}{4}at^2 \bigg) \\ t' & = \frac{-(u)\pm \sqrt{(u)^2-4\big( \frac{1}{2}a\big)\big( -\frac{1}{2}ut-\frac{1}{4}at^2\big)}}{2\big(\frac{1}{2}a\big)} \\ t' & = \frac{-u\pm \sqrt{u^2+aut+\frac{1}{2}a^2t^2}}{a} \\ \dots\textrm{ because } &v-u=at\enspace \dots \\ t' & = \frac{-u\pm \sqrt{u^2+u(v-u)+\frac{1}{2}(v-u)^2}}{a} \\ & = \frac{-u\pm \sqrt{\displaystyle{\frac{u^2+v^2}{2}}}}{a} \end{aligned}$

At mid-point $Y$ the speed $w$ should therefore be

$\begin{aligned} w & =u+at' \\ & = u + a \Bigg( \frac{-u\pm \sqrt{\frac{u^2+v^2}{2}}}{a} \Bigg) \\ & = \pm \sqrt{\displaystyle{\frac{u^2+v^2}{2}}} \end{aligned}$

And the answer is A.

Remark.

What above was done brutally and mechanically, without a sense of physical meaning.

July 28, 2019October 13, 2022

201907280623 Solution to 1982-AL-PHY-I-7

(Non-relativistic approach.)

Set up a 2-D Cartesian coordinate system, the origin being in the position of body $X$ at time $t=0$ , and at the point $(6,0)$ there being body $Y$ .

Then the position of body $X$ and of body $Y$ can each be given by a function of time $t$ :

$\begin{aligned} \mathbf{r}_X(t) & = 3t\, \hat{\mathbf{i}} \\ \mathbf{r}_Y(t) & =6\, \hat{\mathbf{i}} + 4t\, \hat{\mathbf{j}} \end{aligned}$

where $t\in [0,\infty )$ .

The separation $\mathbf{r}_{YX}$ of body $Y$ from body $X$ by time $t$ is:

$\mathbf{r}_{YX}(t)= \mathbf{r}_Y - \mathbf{r}_X =(6-3t)\, \hat{\mathbf{i}} + (4t)\, \hat{\mathbf{j}}$ .

The velocity $\mathbf{v}_{YX}$ of body $Y$ from body $X$ is:

$\begin{aligned} \mathbf{v}_{YX} & =\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\Big(\mathbf{r}_{YX}\Big) \\ & = -3\, \hat{\mathbf{i}} + 4\, \hat{\mathbf{j}} \end{aligned}$

The magnitude $v_{YX}$ of the velocity is

$v_{YX}=|\mathbf{v}_{YX}|=\sqrt{(-3)^2+(4)^2}=5\quad (\mathrm{m\, s^{-1}})$ .

And the answer is B.

All above is overkill.

Notice and

March 31, 2019January 19, 2023

201903310628 Solution to 1980-AL-PHY-I-23

Faraday’s law:

$\varepsilon = -N \displaystyle{\frac{\mathrm{d}\Phi}{\mathrm{d}t}}$

where $\varepsilon$ is the e.m.f. induced, $N$ the number of turns in the coil, $\Phi$ the magnetic flux, also ( $\Phi =BA$ ) the product of magnetic flux density $B$ and area $A$ , $t$ the time, and the negative sign due to Lenz’s law.

The magnitude of the e.m.f. induced in the coil is therefore

$\varepsilon = \bigg| -N\displaystyle{\frac{\Delta \Phi}{\Delta t}} \bigg| = \bigg| -N \displaystyle{\frac{\Phi -0}{t}} \bigg| = N\Phi /t$ .

And the answer is C.

March 31, 2019January 19, 2023

201903310609 Solution to 1980-AL-PHY-I-14

By Snell’s law,

$\begin{aligned} n_1\sin\theta_1 & = n_2\sin\theta_2\\ n_1 \sin 45^\circ & = 1 \times \sin 90^\circ \\ n_1 & = \displaystyle{\frac{1}{\sin 45^\circ}} \\ & = \sqrt{2} \\ \end{aligned}$

And the answer is C.

March 31, 2019January 19, 2023

201903310557 Solution to 2004-AL-PHY-II-16

The original electric potential $V_B$ at point $B$ is due to charge $+Q$ at point $A$ :

$V_B =V_{BA} = k \displaystyle{\frac{(+Q)}{(2r)}}=k \displaystyle{\frac{Q}{2r}}$ .

The final electric potential $V_B$ is due to both charges $+Q$ and $-2Q$ at points $A$ and $C$ :

$V_B' = V_{BA} + V_{BC} = k \displaystyle{\frac{Q}{2r}} + k \displaystyle{\frac{(-2Q)}{(r)}}=-3 \bigg( k \displaystyle{\frac{Q}{2r}} \bigg) =-3V_B$ .

And the answer is B.

March 31, 2019January 19, 2023

201903310540 Solution to 1980-AL-PHY-I-2

The original gravitational force $F$ is

$F=G \displaystyle{\frac{(m)(m)}{(2r)^2}}=G \displaystyle{\frac{m^2}{4r^2}}$ .

The final gravitational force $F'$ is

$F'=G \displaystyle{\frac{(m)(m)}{(2r+4r)^2}}=G \displaystyle{\frac{m^2}{36r^2}}=\displaystyle{\frac{1}{9}}\bigg( G \displaystyle{\frac{m^2}{4r^2}} \bigg) = \displaystyle{\frac{1}{9}}F$ .

Thus $F:F'=9:1$ .

And the answer is B.

March 8, 2019January 19, 2023

201903080359 Solution to 2005-AL-PHY-IIA-12

We assume that the charges on a small sphere can be treated as one point charge.

At first, $P$ , $Q$ , and $R$ are of charge $\pm q$ , $\mp q$ , and $0$ .

Then, $P$ and $R$ are put in contact and share their charges evenly. They are now of charge $\pm \displaystyle{\frac{q}{2}}$ . We keep them separated afterwards.

Later on, $R$ and $Q$ are put in contact and share their charges evenly. Each of them is hence of charge:

$\displaystyle{\frac{\bigg( \pm \displaystyle{\frac{q}{2}}\bigg) +(\mp q)}{2}}=\mp\displaystyle{\frac{q}{4}}$ .

We keep them separated afterwards.

The initial magnitude $F$ of electrostatic force between $P$ and $Q$ is given by:

$F=\bigg| \displaystyle{k\frac{(\pm q)(\mp q)}{r^2}} \bigg|=k \displaystyle{\frac{q^2}{r^2}}$

The final magnitude $F'$ of electrostatic force between them is related to the initial $F$ by:

$F'=\bigg| \displaystyle{k\frac{ ( \pm \frac{q}{2}) (\mp \frac{q}{4})}{r^2}} \bigg|=\displaystyle{\frac{1}{8}}\bigg( k \displaystyle{\frac{q^2}{r^2}} \bigg) = \displaystyle{\frac{1}{8}}F$

$\begin{aligned} \quad & \quad & \textrm{Initially} & \quad & \rightarrow & \quad & P\textrm{ and }R\textrm{ in touch} &\quad & \rightarrow & \quad & R\textrm{ and }Q\textrm{ in touch}\\ P & & \pm q & & & & \pm \displaystyle{\frac{q}{2}} & & & & \boxed{\pm \displaystyle{\frac{q}{2}}} \\ Q & & \mp q & & & & \boxed{\mp q} & & & & \mp \displaystyle{\frac{q}{4}}\\ R & & 0 & & & & \pm \displaystyle{\frac{q}{2}} & & & & \mp \displaystyle{\frac{q}{4}} \end{aligned}$

And the answer is B.