How is the refractive index of water calculated in rainbow formation?

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Homework Help Overview

The discussion revolves around the calculation of the refractive index of water in the context of rainbow formation, with specific reference to the angles of incidence and refraction as illustrated in diagrams related to the path of light through raindrops.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to label angles of incidence and refraction based on a diagram, seeking confirmation of their accuracy. They also question their measurements and calculations of the refractive index, noting discrepancies in their results.

Discussion Status

Participants are engaged in clarifying the correct positioning of the normal line relative to the water's surface and its implications for angle measurement. Some guidance has been offered regarding the definition of the normal line on a spherical surface, and there appears to be a productive exchange of ideas regarding the correct approach to the problem.

Contextual Notes

There is an emphasis on the accuracy of angle measurements and the proper drawing of the normal line, which may be influencing the calculations of the refractive index. The original poster's confusion regarding their results suggests a need for careful consideration of these geometric principles.

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Homework Statement


The diagrams below illustrate the formation of a rainbow. Figure 1 shows the general arrangement and Figure 2 shows the path of a ray through a raindrop with the centre of the raindrop is labelled O.

Figure 1:
Raindrop1.jpg


Figure 2:
Raindrop2.jpg


a.)
Where the ray enters the raindrop in Figure 2, mark the angle of incidence i and the angle of refraction, r.

b.)
Figure 2 is drawn to scale. By taking suitable measurements, show that the refractive index of water is about 1.3.


Homework Equations


[itex]\mu = \frac{\sin i}{\sin r}[/itex]


The Attempt at a Solution


a.)
Raindrop2Attempt.jpg

^ Have I labelled it correct?

b.)
I measured the angle that I have labelled above and got [itex]i = 25^\circ[/itex] and [itex]r = 35^\circ[/itex] and this gives [itex]\mu = 0.74[/itex]. This is incorrect. The correct answer is [itex]i = 53, \ r = 39, \ \implies \mu = 1.27[/itex]. Where have I gone wrong here?
 
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The normal should be perpendicular to the water's surface.
 
So, is it like this?

Raindrop2Attempt2.jpg
 
Or, is it like this? :confused:

Raindrop2Attempt3.jpg
 
You just seem to be drawing random lines lol...

The normal line on a spherical surface at some point P is always the line that goes from P through the center of the circle.

In other words; if you draw a straight line through the center of a circle, then at the points where it intersects the circle the angle between the line and the circle (at that point) is exactly 90 degrees, or perpendicular.
 
Nick89 said:
You just seem to be drawing random lines lol...

The normal line on a spherical surface at some point P is always the line that goes from P through the center of the circle.

In other words; if you draw a straight line through the center of a circle, then at the points where it intersects the circle the angle between the line and the circle (at that point) is exactly 90 degrees, or perpendicular.

Oooh, I understand now. I guess it is this:

Raindrop2Attempt4.jpg


Am I correct?
 
Last edited:
Yes, that looks good.
 

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