Read the following passage and answer the questions below - Leaving Cert Physics - Question 11 - 2019

The pupil's function is to allow light to enter the eye, regulating the amount of light that reaches the retina.

The eye focuses light through a process called accommodation. When focusing on nearby objects, the ciliary muscles contract, causing the lens to become thicker, which increases its refractive power. Conversely, when focusing on distant objects, the ciliary muscles relax, allowing the lens to thin, decreasing its refractive power.

The two most common eye defects are myopia (short sightedness) and hyperopia (long sightedness).

Short sightedness is corrected with a diverging lens, which spreads light rays apart. A sketch of a diverging lens can be illustrated as follows:

     |  |
     |  |  -- Diverging Lens
    _|__|_
   /      \
  /        \
 /          \

The arrows indicate the diverging light rays.

The ray diagram should illustrate the convergence of light rays through a lens to form a real image. The essential components to include are the lens, the object, the focal points, and the resulting image on the opposite side of the lens.

The power required for clear vision is 38 m–1, and the individual has a maximum power of 32 m–1. The power of the lens required is given by:

$P = 38 ext{ m}^{-1} - 32 ext{ m}^{-1} = 6 ext{ m}^{-1}$

The focal length (f) of the lens can be calculated using the formula:

$P = \frac{1}{f}$

Rearranging gives:

$f = \frac{1}{P} = \frac{1}{6 ext{ m}^{-1}} \approx 0.167 ext{ m}$

So, the focal length is approximately 16.7 cm.

Refraction is the bending of light rays as they pass from one medium to another with different densities. This change in speed as light passes through the media causes it to change direction.