# Finding speed of light through medium, when given critical angle of refraction

• seanmcgowan
In summary, the water had something to do with the speed of light in calcite. The equation was rearranged to solve for v. v= (3x10^8)/68.4= 4385964.91228. This value is correct.
seanmcgowan

## Homework Statement

The critical angle of refraction for calcite is 68.4 (degrees) when it forms a boundary with water. Use this information to determine the speed of light in calcite.

## Homework Equations

n=c/v
n= Index of refraction
c= Speed of light in a vacuum
v= Speed of light in medium

## The Attempt at a Solution

I am looking for v so i rearranged the equation to look like this: v=c/n

v= (3x10^8)/68.4= 4385964.91228

Is this right? Did the water have something to do with the equation? If not then could you please help, I have no idea how to work this problem otherwise.

First, and perhaps most significantly, 68.4 degrees is an angle... not an index of refraction (which you have plugged it in as). You also should look up what "critical angle" means... since you will need to use this angle as one of the quantities in your calculation... a calculation that will probably first yield index (which you can then use in the equation you've shown). Hint: also look up information about "Snell's Law"... this is a refraction problem after all, and critical angle has to do with a critical point in the math of Snell's law.

So here's what i did:

using: sin(critical angle)= index of refraction for second medium/ Index of refraction for first medium...

sin(68.4)= 1.333/x
x=1.33/sin(68.4)= 1.44

then using the the n=c/v equation:

1.44= 3x10^8/x
x=3x10^8/1.44= 208333333.33

so would the answer be 2.08x10^8? also, i didn't find any use for snell's law. it would have given me the angle of incedence, which i believe is irrelevant, but i may be wrong.

using the previous eq. u must find refractive index which is sin(angle of incidence)/sin(angle of refraction) then use the refractive index to find the speed of light in another medium by doing this:
v=c/n
where c=3x10^8
and n is the refractive index you found earlier.
Hope this helps

Your attempt at a solution is on the right track. However, there are a few things to consider in order to find the speed of light in calcite.

Firstly, the equation n=c/v is used to relate the speed of light in a vacuum (c) to the speed of light in a medium (v). This means that n is the index of refraction of the medium, which is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. In other words, n=c/v.

In this case, we are given the critical angle of refraction for calcite when it forms a boundary with water. This means that the light is traveling from water into calcite, and the critical angle is the angle at which the light will be refracted such that it travels along the boundary between the two materials. This angle is related to the index of refraction through the equation n=sin(i)/sin(r), where i is the angle of incidence (in this case, the critical angle) and r is the angle of refraction.

So, using the given critical angle of 68.4 degrees, we can find the index of refraction of calcite: n=sin(68.4)/sin(90)= 0.927.

Now, we can use this value of n in the equation v=c/n to find the speed of light in calcite: v=(3x10^8)/(0.927)= 323,877,551.0204 m/s.

Note that the water does not have a direct effect on the equation, as we are only considering the speed of light in calcite. However, the fact that the light is traveling through water before entering calcite does affect the angle of refraction and therefore the index of refraction of calcite.

I hope this helps clarify the problem for you. If you have any further questions, please let me know.

## 1. What is the critical angle of refraction?

The critical angle of refraction is the angle at which light passing through a medium is refracted to an angle of 90 degrees, resulting in total internal reflection. This angle depends on the properties of the medium, such as its refractive index.

## 2. How is the speed of light through a medium related to the critical angle of refraction?

The speed of light through a medium is inversely proportional to the critical angle of refraction. This means that as the critical angle increases, the speed of light through the medium decreases.

## 3. Can the critical angle of refraction be calculated if the speed of light through a medium is known?

Yes, the critical angle of refraction can be calculated using the formula: critical angle = arcsin(1/n), where n is the refractive index of the medium. This formula assumes that the light is passing from a medium with a higher refractive index to one with a lower refractive index.

## 4. How does the critical angle of refraction affect the behavior of light passing through a medium?

If the angle of incidence is less than the critical angle, the light will be refracted as it passes through the medium. However, if the angle of incidence is equal to or greater than the critical angle, the light will be reflected back into the same medium, resulting in total internal reflection.

## 5. What are some real-world applications of knowing the critical angle of refraction?

Understanding the critical angle of refraction is important in a variety of fields, such as optics, telecommunications, and medicine. For example, total internal reflection is used in fiber optic cables to transmit light signals without loss of intensity, and in endoscopes to view internal organs without invasive procedures. Additionally, knowing the critical angle can help determine the optimal angle for designing lenses and mirrors in optical devices.

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