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2020 WAEC Physics Practical You are provided with a glass block, plane mirror, and optical pins.  (i) Place the...

Physics
WAEC 2020

You are provided with a glass block, plane mirror, and optical pins. 

(i) Place the glass block on a drawing sheet and trace its outline ABCD as shown in the diagram above.

(ii) Remove the block, measure and record the width W of the block.

(iii) Draw a normal ON to DC at a point about one-quarter the length of DC.

(iv) Draw a line making an angle i = 10° with the normal.

(v) Replace the block on its outline and mount the plane mirror vertically behind the block such that it makes good contact with the face AB.

(vi) Stick two pins P\(_{1}\) and P\(_{2}\) on the line MO.

(vii) Looking through the face CD, stick two other pins P\(_{3}\) and P\(_{4}\) such that they appear to be in a straight line with the images of pins P\(_{1}\) and P\(_{2}\) seen through the block.

(viii) Join P\(_{3}\) and P\(_{4}\) with a straight line and extend it to touch the face CD at O\(^{1}\).

(ix)Draw a perpendicular line from the midpoint of OO\(^{1}\) to meet AB at Q.

(x) Draw lines OQ, O\(^{1}\)Q and normal O\(^{1}\)N\(^{1}\) produced.

(xi) Measure and record \cos\theta, e, and d.

(xii) Evaluate m = sin e, and n cos\(\frac{\theta}{2}\)

(xii)Repeat the procedure for i = 20°, 30°, 40° and 50.

(xiv) Tabulate your readings.

(xv) Plot a graph with m on the vertical axis and n on the horizontal axis.

(xvi) Determine the slope, s, of the graph and evaluate \cos\theta = 2Ws.

(xvii) State two precautions are taken to ensure accurate results.

(xviii) Sketch a diagram to show the path of the ray through the glass block when the angle of incidence i = 90° in the experiment above.

(xix) A coin lies at the bottom of a tank containing water to a depth of 130cm. If the refractive index of water is 1.3, calculate the apparent displacement of the coin when viewed vertically from above.

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Explanation
i/° \theta/° e/° d/cm m = sine/° n = cos\(\frac{\theta}{2}\)/°
10
20
30
40
50
10.4
19.0
20.0
30.0
30.5
10.0
20.4
30.0
40.0
50.0
3.00
3.90
6.00
7.00
7.50
0.174
0.349
0.500
0.643
0.766
0.996
0.986
0.985
0.966
0.965

Slope = \(\frac{0.766 - 0.174}{0.965 - 0.996}\)

= \(\frac{0.592}{-0.031}\)

= - 19.097°

q = 2ws
q = 2 x 6.5 x -19.097
q = 1248.261
q = -248.3

Open Photo

Precautions:- Neat traces, Pins located vertically, Pins reasonably spaced, Avoided parallax error while reading the protractor/ ruler, zero error avoided on the metre rule.

(b) Refractive index is the ratio of the velocity of light in air to the velocity of light in a medium when light waves pass from air to a material medium. Mathematically,

n = \(\lambda_{1}\) 
      \(\lambda_{2}\). Where; 

\(\lambda_{1}\) = wavelength in air
\(\lambda_{2}\) = wavelength in material
n = refractive index of the material

(ii) The light must be traveling from a denser medium to a less dense medium. The angle of incidence in the denser medium must be greater than the critical angle.


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NECO June/July 2024 - Get offline past questions & answers - Download objective & theory, all in one app 48789
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