HYSICS PRACTICAL SOLUTIONS QUESTION 2: (a.) Measure and record the thickness of the glass block provided. Trace the outline ABCD of the glass block on the sheet of paper as shown. Remove the block and draw Normal at N. Draw an incident ray such that the angle of incidence,i=25degree. Fix two pins at points P and Q on the incident ray. Replace the glass block and fix two other pins at point R and Y such that the pins appear to be in a straight line with the images of the pins at P and Q when viewed through the side DC of the glass block. Remove the block and join the points at R and Y producing the line to meet DC and X. Join NX and measure its length L. Evaluate L^-2 and Sin^2i. Repeat the experiment for i=35degree, 45degree,55degree and 65degree. In each case determine the corresponding values of L, L^-2 and Sin^i. Tabulate your readings. Plot a graph of L^- on the vertical axis and Sin^2i on the horizontal axis, starting both axes from the origin. Determine the slope(s) of the graph and the intercept(I) on the vertical axis. Evaluate the expression: K=(I/S)^1/2. State two precautions taken to ensure accurate results. (i.) Using your graph deduce the value of L when i=0degree. (ii.) State Snells law of refraction and explain why refraction occurs at the boundary between two media. ANSWER: Table of values/Observation: Width of glass block b=6.5cm. S/N: 1,2,3,4,5. P(degree): 25.00,35.00,45.00,55.00,65.00. L(cm): 6.80,7.00,7.30,7.80,8.20. L^2(cm^2): 46.24,49.00,53.29,60.84,67.24. L^-2(cm^-2): 0.0216,0.0204,0.0188,0.0164,0.0149. SinPdegree: 0.04226,0.5736,0.7071,0.8191,0.9063. Sin^2Pdegree: 0.1786,0.3290,0.5000,0.6710,0.8214. NOTE THE FOLLOWINGS: *Draw a table for the above values. *Note that ^ means Raise to Power. Eg: 2^-1 means Two raise to power minus One. *Note that COMMA(,) in the above table means NEXT LINE. Eg: 2,3,4,5 means dat 3 is under 2, 4 is under 3 and 5 is under 4 in a table form. Therefore, S/N has 1,2,3,4,5 under it, in a table as shown in the table of values above. i.e 1 is under S/N and 2 is under 1 and 3 is under 2 and 4 is under 3 and so on..... **************************** Slope= DL^-2(cm^-2)/DSin^2i = 0.0136-0.0224/0.96-0.1 = 0.0088/0.86 = -0.01023(cm^-2) K=(1/5)^1/2 = (0.234/0.01023) ^1/2 = 1.512cm, =1.5cm. Deduction intercept on vertical axis = 0.0234(cm2). PRECAUTIONS: (i.) I avoided error of parallax. (ii.) I ensured that the object pins and the image pins were erect and in a straight line. (bi) When i = OL^-2 = 0.0236cm^-2 = L^-2 = 42.37cm^2. Hence L=6.5cm.
Posted on: Thu, 27 Mar 2014 22:36:03 +0000
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