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2015 WAEC Physics Practical You are provided with a triangular glass prism, four optical pins, and other necessary materials....

Physics
WAEC 2015

You are provided with a triangular glass prism, four optical pins, and other necessary materials.

  • Place the triangular glass prism on a drawing paper and draw its outline UMR. Remove the prism.
  • Measure and record the value of the angle at U.
  • Draw a normal to the line UM at N. Also, draw another line TN to the normal such that \(\phi\)-60. Fix two pins at P\(_{1}\) and P\(_{2}\).
  • Replace the prism and fix two other pains at P\(_{3}\), and P\(_{4}\), such that the pins appear to be in a straight line with the images of the pins at P\(_{1}\) and P\(_{2}\), when viewed from the side UR. Remove the prism.
  • Join points P\(_{3}\), and P\(_{4}\), producing the line to meet TN produced at Z. Draw the normal XY.
  • Measure and record the angle of emergence e and that of deviation d.
  • Repeat the experiment with \(\phi\) = 55°, 50°, 40° and 35°.
  • In each case, measure and record the Corresponding values of e and d.
  • Tabulate your readings.
  • Plot a graph with d on the vertical axis and e on the horizontal axis starting both axes from the origin (0.0) Join your points with a smooth curve.
  • From your graph, obtain the minimum deviation d\(_{m}\) and the corresponding angle of emergence e\(_{m}\) . Hence, calculate the refractive index n of the prism using the formula:

n = \(\frac{sin (\frac{d_{m}+U}{2})}{sin{(\frac{u}{2})}}\)

  • State two precautions taken to obtain accurate results. Attach your traces to your answer booklet.

(b)i. State the conditions necessary for total internal reflection of light to occur.

ii. The critical angle for a transparent substance is 39°. Calculate the refractive index of the substance.

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Explanation

U  = 60°
 

S/N \(\phi\)/° e/° d/°
1 60.0 30.00 2.000
2 55.0 35.00 10.000
3 50.0 40.00 21.000
4 40.0 50.00 26.000
5 35.0 55.00 30.00

Slope = \(\frac{57-30}{38.2} = \frac{27}{33.5}\)

xi. u = 60°
dm = 25°

n = [\(\frac{25° = 60°}{2}\)]

Sin[\(\frac{60°}{2}\)]

n = \(\frac{sin(\frac{85}{2})}{sin30}\)

n = \(\frac{sin47.5}{sin30°} = \frac{0.7373}{0.5000}\)

n = 1.5

PRECAUTIONS;

- ensured neat traces/sharp pencil (Shown)
- ensured that pins were reasonably Spaced.
-  avoided parallax error in reading protractor.

(b)i. - Light must travel from an optically dense medium to optically less dense medium.
- The critical angle must be exceeded. 

ii. n = \(\frac{1}{Sinc}\)

n = \(\frac{1}{Sin30} = \frac{1}{0.6293}\) = 1.59

see graph above


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