2025 HCI J2 BT Q6

2025 HCI J2 BT Q6

13 marks
Block Test

When a light ray passes from a transparent medium to air, it is deflected through an angle. A light ray is shone on a transparent object \(\mathrm{P}\) in the shape of a rectangular prism. The light ray hits the base of the object at point \(A\left( 2,2,-3 \right)\) relative to origin \(O\) and travels in the direction parallel to \(4\mathbf{i}+\mathbf{k}\).

It leaves the top of the object at point \(B\) and the deflected ray passes through point \(C\left( 2,-5,1 \right)\).

The top of the prism is part of a plane with normal vector \(\mathbf{n}\) modelled by the equation \(\mathbf{r}\cdot \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right)=6\).

The diagram above shows a cross-section of the object \(\mathrm{P}\) in the plane of the light ray and \(\mathbf{n}\). The angle, \(\theta \), between the light ray and a line perpendicular to the surface of the object \(\mathrm{P}\) is known as the angle of incidence.

  1. Find the coordinates of point \(B\).[3]
  2. Determine the thickness of the glass object.[4]

A phenomenon known as total internal reflection occurs when the light ray does not leave the transparent medium and is instead reflected within the transparent medium. When object \(\mathrm{P}\) is replaced by another transparent object \(\mathrm{Q}\) with the same dimensions as object \(\mathrm{P}\), total internal reflection occurs. The reflected ray makes the same angle \(\theta \) with the line perpendicular to the surface of object \(\mathrm{Q}\) at the point \(B\) and exits object \(\mathrm{Q}\) at point \(A'\).

  1. Find the coordinates of \(A'\).[4]
  2. Hence, or otherwise, find the vector equation of the reflected ray.[2]

Video Solution:

Video Solution

Video solution locked

A free account saves your progress; this video needs a course plan.

Solution:

Solution locked

680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs a course plan.

Answer:(a) \(\left(6,2,-2\right)\) (b) \(\dfrac{5}{\sqrt3}=\dfrac{5\sqrt3}{3}\text{ units}\) (c) \(\left(\dfrac{20}{3},-\dfrac43,-\dfrac{13}{3}\right)\) (d) \(\mathbf r=\begin{pmatrix}6\\2\\-2\end{pmatrix}+t\begin{pmatrix}2\\-10\\-7\end{pmatrix}\), \(t\geq0\)

Need help? Join our JC Math tuition classes.

Learn more