Answers to Optics Practical 1

Question 1 §

Practical 1 Optics Diagram 1.excalidraw The diagram above represents the question. Using the law of reflection you can calculate that ${\theta }_1={\theta }_1^{\prime },{\theta }_2={\theta }_2^{\prime },{\theta }_3=90^{\circ }-{\theta }_2={\theta }_4$.

Question 2 §

Optics Practical 1 Diagram 2.excalidraw Because the densities are such that $n_1<n_2<n_3⟹{\theta }_1>{\theta }_2>{\theta }_3$

Each ray reflects at each boundary at the same angle as incidence, then Snell’s Law determines the relationship between angles within each material:

$$n_1\sin {\theta }_1=n_2\sin {\theta }_2=n_3\sin {\theta }_3$$

Thus it’s proven also that the presence of water is irrelevant to the angle within the glass.

Question 3 §

The refractive index of air is $n_2=1.00$, and water is $n_1=1.33$. Using this, we can calculate the critical angle using the formula $\sin ({\theta }_c)=\frac{n_1}{n_2}⟹{\theta }_c={\sin }^{-1}\frac{1.33}{1.52}=61.04^{\circ }$ for the glass-water interface. For the water-air interface, it’d be $\sin ({\theta }_c)=\frac{n_2}{n_3}⟹{\theta }_c={\sin }^{-1}\frac{1.00}{1.33}=48.75^{\circ }$, give that the refractive index of crown glass is $n_3=1.52$.

Question 4 §

If modelled as a single ray of light, then at the critical angle total internal reflection will occur and you will not see the light. If modelled realistically as multiple rays of light, this is improbable as each ray will be at a different angle.