Applied Mathematics - Finals Revision Sessions (1-2)

Revision Plan §

Final exam made of two parts: A (shorter methodical questions, solutions required for all questions), and B (longer conceptual questions, highest marking solution used for final marks).

In-class revision: 1 hour optics, 2 hours mechanics, 1 hour waves.

• Practical Solutions
  • Week 1, Problem 2 (Optics)
  • Week 2, Problem 4 (Optics)
  • Week 3, Problem 3 (Optics)
• Revision Questions
  • Kinematics (Mechanics)
  • …

Optics §

Generally, in optics, you should draw a detailed graphical solution to the question, as well as detailed supporting algebraic solution.

Week 1, Problem 2 §

Question §

A layer of water is on the top of a horizontal rectangular slab of glass. There is no wind or other perturbances on the surface of water. A light ray hits the water at some non-zero angle with the normal to the water surface. Draw all resulting rays. Choose your own angle of incidence for the first ray (you don’t need necessarily to choose a numerical value, you can choose a graphical representation) and find angles of travel for all other rays. The indexes of refraction of air, water and glass are: $n_1$, $n_2$, and $n_3$ correspondingly.

Solution §

Assuming that the glass is on a non-reflective surface, let $n_1$ be the refractive index of air, $n_2$ the refractive index of water, and $n_3$ the refractive index of glass, such that:

Applied Mathematics - Finals Revision Sessions (1-2) 2024-01-09 11.16.00.excalidraw

$$n_1<n_2<n_3$$

Using Snell’s Law for Refraction, $n_1\sin ({\theta }_1)=n_2\sin ({\theta }_2)$, then we arrive at the following equations:

$$\begin{array}{c}{n}_1\sin ({\theta }_1)=n_2\sin ({\theta }_2)\\ n_2\sin ({\theta }_2)=n_3\sin ({\theta }_3)\end{array}$$

Which we can then either solve for ${\theta }_1,{\theta }_2,{\theta }_3$, or arrive at the conclusion that $n_1\sin ({\theta }_1)=n_3\sin ({\theta }_3)$, such that the angle of the refracted ray in the glass interface is not dependent on the ray in the previous water interface.

$$\boxed{The\; refracted\; ray\; in\; the\; final\; interface\; does\; not\; depend\; on\; the\; intermediary\; interface.}$$

However, there could also be more rays, due to factors such as Total Internal Reflection, etc. - there could (mathematically) be an infinite number of rays, hence we could not draw them all.

Week 2, Problem 4 §

Question §

A coin of 1.2 cm in diameter is 4 cm away from a biconvex lens that has a focal length of 12 cm. Find the image of the coin. You can choose an orientation of the coin.

Solution §

Graphically, we can represent the problem as below,

Applied Mathematics - Finals Revision Sessions (1-2) 2024-01-09 11.33.04.excalidraw

Such that the image, $I$, is 6cm away from the biconvex lens, and is 1.5x the diameter of the object: hence has a diameter of 1.8cm.

Algebraically, we can use the mirror question and magnification equation to find the same results:

$$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}⟺\frac{1}{12}=\frac{1}{v}+\frac{1}{4}⟺v=-6$$

$$m=-\frac{v}{u}=-\frac{-6}{4}=1.5$$

$$\boxed{The\; image\; has\; a\; diamter\; of\; 1.8cm,\; 6cm\; away\; from\; the\; biconvex\; lens.}$$

Week 3, Problem 3 §

Question §

Light is shining normally to $n_1-n_2$ interface. The intensity of reflected light is $I=(\frac{n_1-n_2}{n_1+n_2})I_0$, where $I_0$ is the incident intensity, and $n_i$ are the refraction indices for the two media.

Find the intensities of the light transmitted through air-glass interface. Find the intensities of the light transmitted through a slab of glass in air.

Solution §

Light is shining normally to the $n_1-n_2$ interface, thus the angle is $90^{\circ }$. Using the equation given in the question,

$$I={\left(\frac{n_1-n_2}{n_1+n_2}\right)}^2\cdot I_0$$

we can calculate the intensity between an air-glass interface in terms of the incidence intensity, using $n_1=1$ and $n_2=1.5$:

$$I_1={\left(\frac{1-1.5}{1+1.5}\right)}^2\cdot I_0=0.04\cdot I_0$$

$$\boxed{Intensity\; transmitted\; through\; the\; air-glass\; interface\; is\; 0.96x\; the\; initial\; intensity.}$$

We can then continue this equation to find the following intensities whilst the light travels through the slab of glass:

$$I_2={\left(\frac{1.5-1}{1.5+1}\right)}^2\cdot (0.96\cdot I_0)=0.0384\cdot I_0$$

Thus, the remaining intensity out of the slab of glass would be $0.96-0.0384$, therefore:

$$I_{\text{final}}=0.9216\cdot I_{\text{initial}}$$

$$\boxed{Intensity\; transmitted\; through\; the\; glass\; slab\; is\; approximately\; 0.92x\; the\; initial\; intensity.}$$

Mechanics §

Generally, in mechanics, ensure you methodically and clearly explain solutions to each question, justifying everything you do.

Exercise 1 §

Question §

Find the velocity and acceleration vectors of a point object, $M$, moving in a Galilean frame, an inertial frame of reference where Newton’s Laws hold, $R=(O,\hat{i},\hat{j})$ with position vector:

$$\vec{r}(t)=\cos (wt)\hat{i}+\sin (wt)\hat{j}$$

Solution §

By definition,

$$\vec{v}(t)=\dot{\vec{r}}(t)=-w\sin (wt)\hat{i}+w\cos (wt)\hat{j}$$

$$\vec{a}(t)=\dot{\vec{v}}(t)=-w^2\cos (wt)\hat{i}-w^2\sin (wt)\hat{j}$$

Exercise 2 §

Question §

Note: the question is lacking some notation, due to the fact that I have no glasses and cannot read it, whoops.

I also gave up writing the solution because I couldn’t read enough - solution and question to be left for the reader, I guess.

What should be the velocity $\vec{v}$ of a different frame, $R^{\prime }(O,\hat{i},\hat{j})$ relative to the original frame (in Exercise 1), $R$, for the motion of the object $M$ to be observed with constant velocity ${\vec{v}}^{\prime }=v^{\prime }\hat{i}$?

Solution §

Using the Law of Composition of Velocities, $\vec{v}(M|O^{\prime })=\vec{v}(M|O)+\vec{v}(O|O^{\prime })$, we can derive the fact that:

$$\begin{array}{rl}\vec{v}(M|O^{\prime })& =\vec{v}(M|O)+\vec{v}(O|O^{\prime })\\ & =(-w\sin (wt)\hat{i}+w\cos (wt)\hat{j})+()\end{array}$$

…