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

To demonstrate Snell's law, you can perform the following experiment:

Apparatus: Glass/plastic block, ray box, laser, optical pins, and a protractor.

Procedure:

1. Shine a ray of light from the ray box at an angle onto the surface of the glass block.
2. Measure the angle of incidence using the protractor and mark it.
3. Draw the incident ray and the refracted ray and observe the path taken by the light.
4. Use a ruler to measure the angles and record the values for i and r.

Observation/Conclusion: From the measurements, you will observe that the ratio of sin(i) to sin(r) remains constant, thereby confirming Snell's law.

Using Snell's law:

$n = \frac{\sin 36°}{\sin 26°}$

Calculating:

• Calculate ( \sin 36° ) and ( \sin 26° ):
  • ( \sin 36° \approx 0.588 )
  • ( \sin 26° \approx 0.438 )

Therefore:

$n = \frac{0.588}{0.438} \approx 1.34$

Thus, the refractive index of water is approximately 1.34.

In the formula (n = \frac{1}{\sin C}), C stands for the critical angle. It is the angle of incidence at which light is refracted along the boundary between two media, resulting in total internal reflection.

To illustrate the paths of rays of light after they strike a converging lens, draw two lines: one parallel to the principal axis that passes through the lens and converges to the focal point, and another line passing through the center of the lens, continuing straight through.

A 'magnified virtual image' refers to an image that appears larger than the actual object and cannot be projected onto a screen. It is formed by diverging light rays that appear to come from a point behind the lens.

To calculate the image distance using the lens formula:

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

Given:

• Focal length, f = 15 cm
• Object distance, u = -20 cm (negative as per convention)

Substituting values:

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

Solving for v: $\frac{1}{v} = \frac{1}{15} + \frac{1}{20}$

Finding a common denominator (60): $\frac{1}{v} = \frac{4}{60} + \frac{3}{60} = \frac{7}{60}$

Thus: $v = \frac{60}{7} \approx 8.57 cm$

Magnification (m) is given by the formula:

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

Substituting the values:

• v = 8.57 cm
• u = -20 cm

Calculating:

$m = -\frac{8.57}{-20} = 0.4285$

So, the magnification is approximately 0.43.

Lenses are used in a variety of applications such as magnifying glasses, eyeglasses, cameras, telescopes, and microscopes.