Light undergoes refraction as shown in the picture - Leaving Cert Physics - Question 7 - 2020

To demonstrate Snell's law, set up a ray box and direct a beam of light at a glass or plastic block at an angle. Use a protractor to measure the angles of incidence and refraction. Record the values, calculate the ratio of the sines of these angles, and confirm that it remains constant.

Using Snell's law:

$n = \frac{\sin i}{\sin r}$

Substituting in the given angles:

$n = \frac{\sin(36°)}{\sin(26°)}$

Calculating the sines:

$n = \frac{0.5878}{0.4384} \approx 1.34$

Thus, the refractive index of water is approximately 1.34.

In the formula n = \frac{1}{\sin C}, C represents the critical angle at which total internal reflection occurs.

To complete the diagram, draw two incident rays towards the converging lens; one parallel to the principal axis that refracts through the focal point on the opposite side, and another ray passing through the focal point on the object side, refracting parallel to the axis after passing through the lens.

A magnified virtual image is formed by a converging lens when the object is placed within the focal length. This type of image cannot be projected onto a screen but can be seen by looking through the lens.

Using the lens formula:

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

where f = 15 cm and u = -20 cm (the object distance is negative), rearranging gives:

$v = \frac{1}{\frac{1}{f} + \frac{1}{u}} = \frac{1}{\frac{1}{15} + \frac{1}{-20}}$

Calculating gives:

$v = \frac{1}{\frac{4 - 3}{60}} = 60 \text{ cm}$

Thus, the image distance is 60 cm.

Magnification (m) is calculated using the formula:

$m = -\frac{v}{u}$

Substituting the values:

$m = -\frac{60}{-20} = 3$

The magnification is thus 3, indicating the image is three times larger than the object.

Lenses are used in magnifying glasses, eyeglasses, cameras, telescopes, etc.