An experiment about the effect of sugar concentration in the index of refraction. I used a tank ,water, laser... I have changed the different amount of concentration and I measured the angle.
My question is:
I Drew two vertical lines. One starting from the point where the laser touches the water surface and the other where the laser touches the bottom of the tank. The a third vertical vertical line connect the two points together into a horizontal line. With this I have an adjacent and a opposite sides of the right-angle triangle. I can calculate the hypotenuse using SOH-CAH-TOA principle. I could calculate the incident angle.
Is this okay for calculating the angle of refraction? I measured it 3 times for every concentration to be sure I would be precise as possible. The other problem is that, my angle of incident is not constant as the volume of water increased with 50% concentration. Then what is the effect of it, is not a controlled variable anymore?
The higher the sugar concentration, the more light bends (higher refractive index). The refractive index is increasing because the solution is getting thicker creating a denser medium with a higher refractive index
The Index of Refraction is defined by Snell's Law which is:
Where 'n' is the index of refraction, and theta (θ) is the angle from the normal.Light will bend through a change in medium, as well as reflect. θ1 and θ2 represent the index of refraction.
The sugar does not dissolve at a constant gradient. There are differences in each direction, due to reasons including uneven sugar distribution, a difference of temperature outside of the water, and convection outside of the tank. It is therefore difficult to tell what the index of refraction is at any given point. The change in angle would need to be looked at, and applied to Snell's Law.
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