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.
Verified Correct Answer: .
Physical Principles & Numerical Solution
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
Source: Myschool Past Questions Archive (Record #60113)