# Refractive index by tracing light rays

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Introduction

Physics Laboratory Report

Refractive index by tracing light rays

Date of experiment: 24/3/2009

Aim of experiment:

The objective of this experiment is:

- to find out the refractive index of a glass block by Snell’s Law
- to find out the refractive index of water by Snell’s Law

Theory:

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 r

As nair = 1, sin i / sin r = nglass

where 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 r

By substituting r =90o, nwater = nglass sin i

where 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 1

Pins x 4

Drawing board x 1

Protractor x 1

A4 paper x 2

Middle

Ray 1 | Ray 2 | Ray 3 | Ray 4 | Ray 5 | |

i (±0.5o) | 25o | 35 o | 45 o | 55 o | 65 o |

r (±0.5o) | 17 o | 23 o | 27 o | 34 o | 36 o |

sin i (±8.73x10-3) | 0.423 | 0.574 | 0.707 | 0.819 | 0.906 |

sin r (±8.73x10-3) | 0.292 | 0.391 | 0.454 | 0.559 | 0.588 |

Mean of sin i = (0.423 + 0.574 + 0.707 + 0.819 + 0.906) / 5 = 0.6858 ± 8.73 x 10-3

Mean of sin r = (0.292 + 0.391 + 0.454 + 0.559 + 0.588) / 5 = 0.4568 ± 8.73 x 10-3

∴ Centroid = (0.4568, 0.6858)

Slope of the graph (m) = sin i / sin r = 1.5777

As nglass = sin i / sin r, ∴ refractive index of the glass block (nglass) = 1.5777

After pivoting the graph about the centroid, the maximum and the minimum slopes are measured.

Maximum slope of the graph (m+) = 1.6839

Minimum slope of the graph (m-) = 1.303

Deviation of maximum slope from slope of best-fit = m+ - m = 1.6839 – 1.5777 = 0.1062

Deviation 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 ± 0.2724

Part B: Refractive index of water by critical angle method

Trial | 1 | 2 | 3 | 4 | 5 |

Critical angle (c)(±0.5o) | 63 o | 62 o | 61 o | 62.5 o | 62 o |

Mean value of c = (63 o + 62 o + 61 o + 62.5 o + 62 o) / 5 = 62.1 o ±0.5o

∴ Critical angle of glass block between water-glass interface = 62.1 o ±0.5o

By nwater = nglass

Conclusion

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.

Conclusion

From 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.

Reference:

Wikipedia

http://en.wikipedia.org/wiki/Water_(data_page)

P.

This student written piece of work is one of many that can be found in our AS and A Level Waves & Cosmology section.

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