Refractive index of glass

Hypothesis:
According to Snell’s Law:

Sin i =n
Sin r

Where,
i = angle of incidence.
r=angle of refraction.
n=refractive index.

Procedure:

1) The glass block was placed on thepaper and its outline was traced onto a sheet of paper.
2) A normal was drawn (perpendicular line)
3) By using a protractor, an incident ray was drawn such that the angle of incidence is 100.4) The sheet of paper was stuck on a piece of cardboard; two pins were aligned on the incoming incident ray and labeled p1 and p2.
5) From the other side of the block, p1 and p2 appeared to bein a straight line and two other pins, p3 and p4, were placed accordingly.
6) The block was removed and lines were drawn joining p1 to p3 and p3 to p4. It was seen that the incident and therefracted rays were not in a straight line.
7) The angle of refraction was recorded.
8) Then the steps 1 to 7 were repeated for 5 more angles of incidence (20o to 70o).
9) The sine of angle ofincidence (sin i) and angle of refraction (sin r) were calculated. A graph of sin i against sin r was plotted to determine the refractive index of glass (gradient of graph).

Diagram:Results:
|Angle of incidence. |Angle of refraction. | | |
|i/º |r/º |sin i|sin r |
|10.0 |5.0 |0.182 |0.091 |
|20.0 |13.0|0.341 |0.223 |
|40.0 |33.0 |0.641 |0.542 |
|50.0|35.0 |0.773 |0.573 |
|60.0 |45.0 |0.871 |0.712…