Aim of experiment:The objective of this experiment is:1. to find out the refractive index of a glass block by Snell’s Law2. to find out the refractive index of water by Snell’s LawTheory:In Part A of this experiment, the angle of incidence (i) and the angle of refraction (r) between air and a glass block were measured.
As stated by Snell’s Law, the following relation can be concluded:nair sin i = nglass sin rAs nair = 1, sin i / sin r = nglasswhere nair and nglass are the refractive indexes of air and the glass block respectively.
Hence, by the above equation and measurements of i and r, the refractive index of the glass block can be determined.In Part B of this experiment, the angle of incidence (i) between water and the glass block were measured.As stated by Snell’s Law, the following relation can be concluded:nglass sin i = nwater sin rBy substituting r =90o, nwater = nglass sin iwhere nwater is the refractive index of water.
Hence, by the above equation, as the refractive index of water is found and i is measured, the refractive index of water can be determined.Apparatus:Rectangular glass block x 1Pins x 4Drawing board x 1Protractor x 1A4 paper x 2Beaker of water x 1Procedures:Part A: Refractive index of a glass block1. A sheet of paper was fixed on the drawing board and the glass block was placed in the middle. The outline of the glass block was sketched by using a sharp pencil.
2. The glass block was removed and 5 straight lines were drawn to represent light rays with angle of incidence varying from 25o to 65o as shown in Fig.1. The lines were labelled as Ray 1 to Ray 5.
For each line, the normal is drawn as dotted line and the angle of incidence (i) was measured.3. The glass block was placed back to the original position.4.
Two drawing pins were placed vertically at any two points along Ray 1.5. The pins were viewed from the opposite side of the glass block and two other pins were placed so that the four pins appeared in line. The path of Ray 1 was completed.
6. The angle of refraction (r) in the glass block was measured and the results were recorded in a table.7. Steps 3 to 6 were repeated with Rays 2 to 5.
8. A graph of sin i against sin r was plotted.Part B: Refractive index of water by critical angle method1. A sheet of paper was fixed on the drawing board and the glass block was placed in the middle.
The outline of the glass block was sketched by using a sharp pencil.2. A small strip of paper of about 1cm wide and of length equal to the thickness of the glass block was cut and a fine dark line was drawn down the middle of the strip. It was smeared with water and stuck on the longer side AD of the glass block.
A vertical protractor was used to check that the line is vertical and the position of the vertical line was marked at P.3. Beginning with my line of sight near C, I looked into side CD so that the dark line could be seen clearly. My line of sight moved towards D.
The line was noticed to disappear suddenly from view at a certain point.4. Two pins were placed to mark the direction beyond which the dark line just could not be seen.5.
The glass block was removed and the marks were joined to meet CD at Q. The ray PQ in glass was completed and the normal at P was drawn as a dotted line.6. The angle between PQ and this normal was measured and this is the critical angle (c) for light between the water-glass interface.
The result was recorded in a table.7. The experiment was repeated with different positions of the strip.Precautions:1.
The glass block should not be moved even for a little when pins have not been all placed for a ray.2. Every pin placed should be ensured to be vertical by viewing it with a vertical protractor from at least two directions.3.
When placing back the glass block after marking light rays, care should be taken that the orientation of the block should be the same as before.4. The dark line marked on the strip in Part B should be marked with a ball pen as the ink would not dissolve much in water.Data and data analysis:Part A: Refractive index of a glass blockRay 1Ray 2Ray 3Ray 4Ray 5i (ï¿½0.
5o)25o35 o45 o55 o65 or (ï¿½0.5o)17 o23 o27 o34 o36 osin i (ï¿½8.73×10-3)0.4230.
5740.7070.8190.906sin r (ï¿½8.
5590.588Mean of sin i = (0.423 + 0.574 + 0.
707 + 0.819 + 0.906) / 5 = 0.6858 ï¿½ 8.
73 x 10-3Mean of sin r = (0.292 + 0.391 + 0.454 + 0.
559 + 0.588) / 5 = 0.4568 ï¿½ 8.73 x 10-3? Centroid = (0.
4568, 0.6858)Slope of the graph (m) = sin i / sin r = 1.5777As nglass = sin i / sin r, ? refractive index of the glass block (nglass) = 1.5777After pivoting the graph about the centroid, the maximum and the minimum slopes are measured.
Maximum slope of the graph (m+) = 1.6839Minimum slope of the graph (m-) = 1.303Deviation of maximum slope from slope of best-fit = m+ – m = 1.6839 – 1.
5777 = 0.1062Deviation of minimum slope from slope of best-fit = m – m- = 1.5777 – 1.303 = 0.
2724? Maximum error in slope = 0.2724? Refractive index of glass (nglass) = 1.5777 ï¿½ 0.2724Part B: Refractive index of water by critical angle methodTrial12345Critical angle (c)(ï¿½0.
5o)63 o62 o61 o62.5 o62 oMean value of c = (63 o + 62 o + 61 o + 62.5 o + 62 o) / 5 = 62.1 o ï¿½0.
5o? Critical angle of glass block between water-glass interface = 62.1 o ï¿½0.5oBy nwater = nglass sin c and the refractive index of the glass block obtained in Part A,nwater = 1.5777 sin 62.
1o ï¿½ 1.39Error Analysis:Part A: Refractive index of a glass blockAbsolute error of angle of incidence and angle of refraction = ï¿½ 0.5oAbsolute error of sin i and sin r = ï¿½ 8.73×10-3Ray 1Ray 2Ray 3Ray 4Ray 5% error in i2.
00 %1.43 %1.11 %0.909 %0.
769 %% error in r2.94 %2.17 %1.85 %1.
47 %1.39 %% error in sin i2.06 %1.52 %1.
23 %1.07 %0.964 %% error in sin r2.99 %2.
23 %1.92 %1.56 %1.48 %Percentage error in refractive index of glass (nglass) = 0.
2724 / 1.5777 x 100% ï¿½ 17.3 %Part B: Refractive index of water by critical angle methodTrial12345% error in c0.794 %0.
806 %0.820 %0.800 %0.806 %Percentage error in critical angle of glass block between water-glass interface= 0.
5o / 62.1o x 100% = 0.805 %According to Wikipedia, the standard refractive index of water is 1.333.
? Percentage error in refractive index of water = (1.39 – 1.333) / 1.333 x 100 % ï¿½ 4.
28 %Sources of error1. In Parts A and B, the pins may not be placed exactly vertical and this will results in an inaccurate observation and measurement of the angles.2. In Part B of the experiment, the disappearance of the line on the strip may not be clearly determined by our naked eyes and this leads to inaccurate results.
3. As the density of the glass block may not be evenly disturbed, there may be a gradual change in refractive index within the whole glass block. Hence, the light ray in the glass block may be bent and this leads to an error in measurement of angles.4.
As the edges of the glass block have been scratched, the image of the line may not be seen clearly because the scratches obscure its image.5. As there are scratches on the surface of the glass block, bending of light due to the scratches may occur when light travel through them, and this will lead to a significant error.Improvement1.
When placing the pins, two observers can be involved in ensuring the pins are vertically placed, with one looking through the glass block and placing the pins at the right position and the other making sure the pins are vertical using a protractor.2. In Part B of the experiment, the position at which the line on the strip just disappears can be determined by several observers, in order to reduce the error in observation and measurement.3.
A glass block of even density should be used to avoid gradual change of refractive index of the glass block, for the sake of preventing the bending of light in the glass block.4. A glass block with smooth surface and without scratches should be used. This reduces the bending and blocking of light by the scratches and hence increasing the accuracy of the experiment.
DiscussionIn this experiment, the paths of light rays are actually traced by our naked eyes, which may seem to be not very accurate. Consequently, it is very natural to think whether this experiment can be modified by using a ray box instead, to replace using pins, and to obtain results of higher accuracy.Nevertheless, there will be difficulties which are difficult to be overcome if we really do so.Firstly, it will be very formidable to measure the angles of incidence and angles of refraction.
As light rays are used, if we remove the glass block and attempt to measure the angles, the bending of light will not occur and hence the measurement cannot be carried out.Secondly, if we attempt to measure the angle from the top of the glass block, the measurement will be very inaccurate as there occurs bending of light from the light rays to our eyes through the glass-air interface.Thirdly, as the beam of light rays has a wide width, there can be much deviation from measuring the angles, leading to a great error in the measurements and results.Lastly, if we attempt to draw lines to mark the paths of light rays, it is also very difficult to do so.
Drawing the lines, the pen, ruler or our hands will block the light and hence we cannot mark the path clearly. Therefore, it is not possible to do so.On balance, there will exist a multitude of difficulties if we make use of a ray box to carry out this experiment. Consequently, it will even be more accurate to use our eyes than to use a ray box in performing this experiment.
ConclusionFrom the experiment, the refractive index of the glass block is 1.5777, with a possible and maximum deviation of 17.3 %. Besides, the refractive index of water is found to be 1.39, which is of 4.28% different from the standard value.
Cite this Refractive index by tracing light rays
Refractive index by tracing light rays. (2017, Aug 23). Retrieved from https://graduateway.com/refractive-index-by-tracing-light-rays/