Using the diagram as a out the guide carry out the following instructions:
(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.
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:
(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
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