Optics Coursework 1

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Question A §

Optics Coursework 1 - Q1.excalidraw The diagram above demonstrates the reflection pattern of the laser pointer by using the law of reflection ${\theta }_1={\theta }_2$.

Question B §

Optics Coursework 1 - Q2.excalidraw Using trigonometry, since each triangle is similar (all with angles $60^{\circ }$ and $30^{\circ }$), it is a simple process of multiplying and dividing by $\sqrt{3}$ (this is the ratio between the sides, $1:\sqrt{3}$.

$⟹c=\frac{a}{4}\cdot \sqrt{3}=\frac{\sqrt{3}}{4}a$ $⟹d=a-c=a-\frac{\sqrt{3}}{4}a=a(1-\frac{\sqrt{3}}{4})$ $⟹e=\frac{d}{\sqrt{3}}=\frac{a(1-\frac{\sqrt{3}}{4})}{\sqrt{3}}=a(\frac{1}{\sqrt{3}}-\frac{1}{4})$ $f=\frac{a}{\sqrt{3}},\therefore \overline{OC}=e+f=a(\frac{1}{\sqrt{3}}-\frac{1}{4})+\frac{a}{\sqrt{3}}$

$$⟹\overline{OC}=a\left(\frac{2}{\sqrt{3}}-\frac{1}{4}\right)=\left(\frac{8-\sqrt{3}}{4\sqrt{3}}\right)\cdot a$$

Approximately, this makes the distance at which $c$ is located from the bottom left corner $O$ about $0.9a$.