TEST OF PRACTICAL KNOWLEDGE QUESTION Using the diagram as a out the guide carry out the following instructions: Place the pin O horizontally inside the cyli...

Answer Details

Test of practical knowledge: real and apparent depth (refraction at a water surface)

The pin O is placed horizontally at the bottom of the cylinder. Water is poured in so that the length of the water column \(l = SO\), where S is the position of the water meniscus. A second (search) pin P in the cork on the retort stand is adjusted vertically until, viewed from above, it coincides with no parallax with the image of O formed by refraction at S. The distance \(h = PO\) is read and recorded. The procedure is repeated for \(l = 15,\ 20,\ 25\) and \(30\ \text{cm}\).

Table of readings

\(l\) (cm)	\(h\) (cm)
10.0	7.50
15.0	11.25
20.0	15.00
25.0	18.75
30.0	22.50

Graph of \(h\) against \(l\)

Plotting \(h\) on the vertical axis against \(l\) on the horizontal axis gives a straight line passing through the origin.

Slope of the graph

Taking two widely separated points on the line of best fit, \((l_1, h_1) = (10.0,\ 7.50)\) and \((l_2, h_2) = (30.0,\ 22.50)\):

\[ s = \frac{\Delta h}{\Delta l} = \frac{22.50 - 7.50}{30.0 - 10.0} = \frac{15.00}{20.00} = 0.625 \]

Evaluation

\[ K_1 = 1 - s = 1 - 0.625 = 0.375 \]\[ K_2 = \frac{1}{K_1} = \frac{1}{0.375} = 0.67 \]

Two precautions

• The lower level of the water meniscus was read with the eye at the same horizontal level as the meniscus to avoid parallax error.
• The zero error of the metre rule was noted and allowed for when measuring \(h\) and \(l\).

(b)(i) Total internal reflection of light

Total internal reflection occurs when light travels from an optically denser medium into a less dense medium and the angle of incidence in the denser medium is greater than the critical angle. Under this condition no light is refracted into the less dense medium; all of it is reflected back into the denser medium, obeying the ordinary laws of reflection.

(b)(ii) Apparent upward displacement of the print

Real depth (thickness of glass) \(D = 6\ \text{cm}\), refractive index \(_a\mu_g = 1.5\).

\[ _a\mu_g = \frac{\text{real depth}}{\text{apparent depth}} \quad\Rightarrow\quad \text{apparent depth} = \frac{6}{1.5} = 4\ \text{cm} \]\[ \text{apparent upward displacement} = \text{real depth} - \text{apparent depth} = 6 - 4 = 2.0\ \text{cm} \]

The print appears raised (displaced upward) by \(2.0\ \text{cm}\).