Using the diagram above as a guide, carry out the following instructions: Fix a plain sheet of paper on the Celotex board. Place the rectangular glass prism...
Question 1 Report
Using the diagram above as a guide, carry out the following instructions:
- Fix a plain sheet of paper on the Celotex board.
- Place the rectangular glass prism on the paper and trace its outline, ABCD. Remove the prism.
- Draw a normal NMP to meet AB and DC at M and P respectively such that \(|AM|=DP=\)2.0cm.
- Trace the ray PQ with two pins, \(P_{1}\) and \(P_{2}\) at P and Q respectively such that angle MPQ = i = 50º
- Replace the prism on its outline. Trace the emergent ray with two other pins \(P_{3}\) and \(P_{4}\) such that they lie in a straight line with \(P_{2}\) and the image of \(P_{1}\) viewed through the glass prism.
- Measure and record \(\theta\), the angle between the emergent ray and face AB of the glass prism.
- Evaluate \(\cos \theta\) and \(\sin\) i.
- Repeat the procedure for four other values of i = 10°, 15°,20° and25°.
- Tabulate your readings.
- Plot a graph with \(\cos \theta\) on the vertical axis and \(\sin\) i on the horizontal axis.
- Determine the slope, s, of the graph
- State two precautions taken to ensure accurate results. Attach your traces to your answer booklet
(b)i. State the laws of refraction of light
ii. Explain what is meant by the statement the refractive index of a material is 1.65
Answer Details
Test of Practical Knowledge: Refraction through a Rectangular Glass Prism
The outline ABCD is traced, the normal NMP drawn with \(|AM| = |DP| = 2.0\,\text{cm}\), and for each angle of incidence \(i\) the emergent ray is located with pins P3 and P4. The angle \(\theta\) between the emergent ray and face AB is measured, and \(\cos\theta\) and \(\sin i\) are computed.
Table of Readings
| \(i\,/\,^{\circ}\) | \(\theta\,/\,^{\circ}\) | \(\cos\theta\) | \(\sin i\) |
|---|---|---|---|
| 5.0 | 83.0 | 0.137 | 0.087 |
| 10.0 | 74.0 | 0.276 | 0.174 |
| 15.0 | 65.0 | 0.423 | 0.259 |
| 20.0 | 56.0 | 0.559 | 0.342 |
| 25.0 | 48.0 | 0.669 | 0.423 |
Graph of \(\cos\theta\) against \(\sin i\)
The points are plotted with \(\cos\theta\) on the vertical axis and \(\sin i\) on the horizontal axis, and a straight line of best fit is drawn through them.
Determination of the Slope
Two widely separated points are read from the line of best fit:
\(A = (\sin i,\ \cos\theta) = (0.150,\ 0.250)\) and \(B = (0.450,\ 0.720)\).
\[ s = \frac{\Delta(\cos\theta)}{\Delta(\sin i)} = \frac{0.720 - 0.250}{0.450 - 0.150} = \frac{0.470}{0.300} = 1.567 \]Slope, \(s = 1.57\).
Two Precautions
- The pins were fixed erect (vertical) and spaced about 4 cm apart so that the ray directions were accurately located.
- Parallax error was avoided by reading the protractor and metre rule with the eye directly above the scale, and sharp, neat traces were made with a well-sharpened pencil.
(b)(i) Laws of Refraction of Light
- First law: The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
- Second law (Snell's law): The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media, \(\dfrac{\sin i}{\sin r} = n\), or \(n_1 \sin i = n_2 \sin r\), where the symbols have their usual meanings.
(b)(ii) Meaning of "the refractive index of a material is 1.65"
It means that the ratio of the speed of light in vacuum (air) to the speed of light in the material is 1.65, i.e. \(n = \dfrac{c}{v} = 1.65\). Equivalently, for light passing from air into the material, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is 1.65, so light travels 1.65 times more slowly in the material and bends towards the normal on entering it.
Download The App On Google Playstore
Everything you need to excel in your exams
