A ray of light can undergo both reflection and refraction - Leaving Cert Physics - Question 7 - 2017

1. The angle of incidence is equal to the angle of reflection.
2. The incident ray, the reflected ray, and the normal at the point of incidence all lie in the same plane.

The periscope consists of two mirrors positioned at 45-degree angles to the light path. When light from an object enters the top of the periscope, it reflects off the first mirror, travels down the tube, and reflects off the second mirror, enabling the observer to see the object at a different level.

The driver can read the word 'AMBULANCE' correctly in the mirror because the image is laterally inverted. This means that the left side of the word appears on the right side in the mirror and vice versa, allowing the driver to recognize the word as it appears in its intended orientation.

The diagram should illustrate a light ray hitting the boundary between two media at an angle greater than the critical angle, leading to total internal reflection. The diagram includes indications of the media, angle of incidence, and the internal reflection path.

The diagram should show light entering one end of the optical fibre, bouncing off the internal walls due to total internal reflection, and continuing through the fibre to the other end. Labels should indicate the light ray, the core, cladding, and the angle of incidence.

To find the angle at which total internal reflection occurs, we can use Snell's Law and the refractive index:

$n = \frac{c}{v}$

For an optical fibre with a refractive index of 1.5, the critical angle $C$ can be calculated using the formula:

$\sin C = \frac{1}{n} = \frac{1}{1.5} = 0.6666$

Thus, the critical angle $C$ is:

$C = \arcsin(0.6666) \approx 41.81^\circ$