Determining the Index of Refraction

[cse63146]

Homework Statement

PART A) What is the frequency of blue light that has a wavelength of 450 nm?

PART B) What is the frequency of red light that has a wavelength of 650 nm?

PART C) What is the index of refraction of a material in which the red-light wavelength is 450 nm?

Homework Equations

$$v = f \lambda$$

The Attempt at a Solution

PART A

$$f = \frac{v}{\lambda} = \frac{3*10^8}{450*10^{-9}} = 4.62*10^{14} Hz$$

*CORRECT*

PART B

$$f = \frac{v}{\lambda} = \frac{3*10^8}{650*10^{-9}} = 6.67*10^{14} Hz$$

*CORRECT*

PART C

Not sure. Can someone point me in the right direction?

 

[negatifzeo]

I want to help but I'm not sure if this is right. The index if refraction is c/v. Maybe you need to use the frequency you got from part b and determine the new velocity of the red light with the new wavelength? Then divide c by this to get the refraction index?

 

[cse63146]

I want to help but I'm not sure if this is right. The index if refraction is c/v. Maybe you need to use the frequency you got from part b and determine the new velocity of the red light with the new wavelength? Then divide c by this to get the refraction index?

is frequency the same for red light regardless of it's wavelength?

 

[negatifzeo]

is frequency the same for red light regardless of it's wavelength?

Not if it's traveling at the speed of light. But my understanding is that refraction slows down light waves, so you might try using the same frequency you had before to find the new speed. Again, I am a student probably around the same place you are in your studies so I don't know. Thats what I would try though

 

[Dick]

negatifzeo is correct. The frequency of the light doesn't change. Only the velocity and hence the wavelength.

 

[cse63146]

Yep, you were correct. I multiplyed the frequncy I got from Part B by the 450 nm to get v, then I just divided it c by it to get the index of redraction.

Thank you.