When a light ray passes from air to glass, it is deflected through an angle. The light ray ABC starts at point A \((1, 2, 2)\), and enters a glass object at point B \((0, 0, 2)\). The surface of the glass object is a plane with normal vector \(\mathbf{n}\). Fig. 7 shows a cross-section of the glass object in the plane of the light ray and \(\mathbf{n}\).
\includegraphics{figure_7}
- Find the vector \(\overrightarrow{AB}\) and a vector equation of the line AB. [2]
The surface of the glass object is a plane with equation \(x + z = 2\). AB makes an acute angle \(\theta\) with the normal to this plane.
- Write down the normal vector \(\mathbf{n}\), and hence calculate \(\theta\), giving your answer in degrees. [5]
The line BC has vector equation \(\mathbf{r} = \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix} + \mu \begin{pmatrix} -2 \\ -2 \\ -1 \end{pmatrix}\). This line makes an acute angle \(\phi\) with the normal to the plane.
- Show that \(\phi = 45°\). [3]
- Snell's Law states that \(\sin\theta = k\sin\phi\), where \(k\) is a constant called the refractive index. Find \(k\). [2]
The light ray leaves the glass object through a plane with equation \(x + z = -1\). Units are centimetres.
- Find the point of intersection of the line BC with the plane \(x + z = -1\). Hence find the distance the light ray travels through the glass object. [5]