QUESTION 6 [14 marks]
QUESTION 6.1
Question 6.1.1 [2 marks]
Define the term angle of incidence.
The angle of incidence is defined as the angle between the incident ray (the ray of light approaching a surface) and the normal (an imaginary line perpendicular to the surface) at the point of incidence.
Question 6.1.2 [3 marks]
Calculate the refractive index of material X using the data in the graph.
The refractive index (n) can be calculated using Snell's Law, which is stated as:
\[
n = \frac{\sin(i)}{\sin(r)}
\]
where \(i\) is the angle of incidence and \(r\) is the angle of refraction. If the values from the graph are, for instance, \(i = 30^\circ\) and \(r = 20^\circ\), we can proceed as follows:
-
Calculate \(\sin(30^\circ)\) and \(\sin(20^\circ)\):
- \(\sin(30^\circ) = 0.5\)
- \(\sin(20^\circ) \approx 0.342\)
-
Substitute into the formula: \[ n = \frac{\sin(30^\circ)}{\sin(20^\circ)} = \frac{0.5}{0.342} \approx 1.46 \]
Assuming you refer to specific values from the graph in your answer, provide those values for precise calculation.
Question 6.1.3 [3 marks]
Calculate the speed of light through material X.
The speed of light in a medium can be calculated using the formula:
\[
v = \frac{c}{n}
\]
where \(c\) is the speed of light in a vacuum (approximately \(3.0 \times 10^8 , \text{m/s}\)) and \(n\) is the refractive index calculated previously. Using the example \(n = 1.46\):
\[ v = \frac{3.0 \times 10^8 , \text{m/s}}{1.46} \approx 2.05 \times 10^8 , \text{m/s} \]
QUESTION 6.2
Question 6.2.1 [4 marks]
Calculate the refractive index of material Y.
Using Snell's Law:
\[
n = \frac{\sin(i)}{\sin(r)}
\]
Where \(i = 7^\circ\) and \(r = 13^\circ\):
-
Calculate \(\sin(7^\circ)\) and \(\sin(13^\circ)\):
- \(\sin(7^\circ) \approx 0.121\)
- \(\sin(13^\circ) \approx 0.225\)
-
Substitute into the formula: \[ n = \frac{\sin(7^\circ)}{\sin(13^\circ)} = \frac{0.121}{0.225} \approx 0.538 \]
Question 6.2.2 [2 marks]
Why is it important to consider the refractive indices of possible contact lenses? Refer to the criteria reflection and refraction when explaining your answer.
The refractive index is crucial in designing contact lenses because it determines how light is bent (refracted) as it passes through the lens and into the eye. The refractive index affects the focal point, influencing vision correction. A proper match between the lens and the eye's refractive index minimizes reflection at the lens surface, reducing glare and enhancing visual clarity. Lenses with appropriate refractive indices ensure effective transmission of light with minimal distortion.
QUESTION 7 [11 marks]
QUESTION 7.1
Question 7.1.1 [2 marks]
Define the term diffraction.
Diffraction is the bending of waves around obstacles and the spreading out of waves when they pass through narrow apertures. It occurs with all types of waves, including light, and is most pronounced when the wavelength of the wave is comparable to the size of the opening or obstacle.
Question 7.1.2 [2 marks]
Complete the pattern of interference for the full section as shown on the screen.
(The actual diagram or data isn't provided. However, students should replicate the given light intensity curve from the figure to illustrate the central bright band and surrounding dark and bright fringes.)
Question 7.1.3 [1 mark]
State ONE safety precaution that a person should take when the above device is used.
One safety precaution is to avoid looking directly into the laser beam to prevent eye injury.
Question 7.1.4 [1 mark]
What type of interference is taking place in the middle of the central bright band?
Constructive interference is taking place in the middle of the central bright band, where the waves reinforce each other, resulting in increased intensity.
Question 7.1.5 [2 marks]
Name TWO changes that the person can make to decrease the distance between the centre of the central bright band and the first dark band.
- Decrease the width of the slit, which increases the diffraction effect.
- Decrease the wavelength of the light source used, which will reduce the separation of the interference patterns.
QUESTION 7.2
Question 7.2.1 [1 mark]
Identify the diffraction pattern, which is produced by the blue laser.
The diffraction pattern produced by the blue laser is the pattern with narrower and more closely spaced fringes, which indicates a shorter wavelength.
Question 7.2.2 [2 marks]
Give a reason for your answer to Question 7.2.1.
The blue laser has a shorter wavelength than the red laser. According to the principles of diffraction, shorter wavelengths result in narrower fringes and closer spacing in the diffraction pattern due to the equation relating fringe spacing to wavelength. Thus, the blue light's pattern is expected to show this characteristic compared to that produced by the red laser.