4 When a light ray passes from air to glass, it is deflected through an angle. The light ray ABC starts at point \(\mathrm { A } ( 1,2,2 )\), and enters a glass object at point \(\mathrm { B } ( 0,0,2 )\). 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 \(\mathbf { n }\).
\begin{figure}[h]
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\caption{Fig. 7}
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- Find the vector \(\overrightarrow { \mathrm { AB } }\) and a vector equation of the line AB .
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.
The line BC has vector equation \(\mathbf { r } = \left( \begin{array} { l } 0
0
2 \end{array} \right) + \mu \left( \begin{array} { l } - 2
- 2
- 1 \end{array} \right)\). This line makes an acute angle \(\phi\) with the
normal to the plane. normal to the plane. - Show that \(\phi = 45 ^ { \circ }\).
- Snell's Law states that \(\sin \theta = k \sin \phi\), where \(k\) is a constant called the refractive index. Find \(k\).
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.