Using the diagram as a out the guide carry out the following instructions:
- Place the pin O horizontally inside the cylinder provided. Pour some water on the pin in the cylinder such that the length of the water column / = SO = 10.0cm, where S place the represents the water meniscus.
- lnsert another pin, P in the cork held by the boss of the retort stand.
- Adjust the position of P vertically upward or 0 formed by refraction at S.
- Read and record the distance h = PO.
- Repeat the procedure for four other values of 1= 15, 20, 25, and 30cm.
- In each case measure and record the corresponding value of h. Tabulate your readings.
- Plot a graph of h on the vertical axis against l on the horizontal axis.
- Determine the slope, s, of the graph
- Evaluate (i) K\(_{1}\) = 1 – S
(ii) K\(_{2}\) = \(\frac{1}{k}\) - State two precautions taken to ensure accurate results.
(b)i. Explain the total internal reflection of light.
ii. A rectangular glass prism of thickness 6 cm and refractive index 1.5 is placed on the page of a book. The prints on the book are viewed vertically down Determine the apparent upward displacement of the print.
Explanation
Table of values
| Icm\(^{3}\) | hcm\(^{3}\) |
| 10.0 | 7.50 |
| 15.0 | 11.25 |
| 20.0 | 15.00 |
| 25.0 | 18.75 |
| 30.0 | 22.50 |
Slope (s) = \(\frac{\bigtriangleup {h}}{\bigtriangleup {|}} = \frac{12.50}{20.00}\) = 0.625
Evaluate: (i) K\(_{1}\) = 1 - S = 1 - 0.625 = 0.375
(ii) K\(_{2}\) = \(\frac{1}{K} = \frac{1}{0.375}\) = 0.67
Precautions:
- The lower level of the meniscus of water is read at eye level.
- Parallax error is avoided in reading the scale to determine h and I
- Zero error of metre rule is noted.
(b)0) Total internal reflection of light occurs when light travels from a dense medium to a less dense medium so that the angle of incidence (in the dense medium) is greater than the critical angle
ii. Let apparent upward displacement of points be t, real depth, D = 6cm. Therefore apparent depth = D - t = 6 - t
Refractive index aug =1.5 = \(\frac{\text {Real depth}}{\text {Apparent depth}}\)
= 1.5 = \(\frac{6}{6 - t}\), (6-t) x 1.5 = 6cm
\(\therefore\) 9-1.5t = 6
i.e t = \(\frac{9-6}{1.5}\) = 2.0cm
OR
Refractive index a\(\mu\)g = \(\frac{\text {real depth}}{\text {apparent depth}}\)
\(\therefore\) Apparent depth = \(\frac{\text {real depth}}{a \mu g}\) = 6 = 4
1.5
Displacement = real depth - apparent depth
= (6 - 4) cm = 2.0cm