Snell law and EM wave properties

Question

1.A light wave has a frequency of  4 x 10⁸. What is the wavelength of the light wave? (Assume the speed of light to be 3 x 10⁸.)

2.A wavelength of light is 300 m. What is the frequency?

3.

Light travels through a material at a speed of 7.5 x 10⁸ m/s. What is the index of refraction for the material?

*answer and explanation please*

Answer 1

Hi there. Let me help you out.

1. First, ponder over how the frequency of a wave is inversely proportional to its wavelength. It is important to recognize which numbers represent which quantities. In this exercise, the frequency of a light wave (which is usually denoted ƒ) is provided along with a kind-of-a-hint that the other quantity you will be using is the speed of light, c. At this point you should search through the EM section of a sheet of relevant equations to find the equation which relates f, c, and wavelength (typically denoted λ). The correct equation to discover is the wave identity relating frequency and wavelength via the velocity of the wave, ƒ = ν/λ. In the case of EM waves, this then looks like ƒ = c/λ. Always keeping track of units, we rearrange terms in order to solve for the wavelength in meters as follows:
2. ƒ = c/λ ⇒ λ = c/ƒ = (3×10⁸ ms^-1) / (4×10⁸ s^-1) = 0.75 m.
3. This second question is the reverse procedure of the first question. Simply plug the numbers into the original equation we used in 1.:
4. ƒ = c/λ = (3×10⁸ ms^-1) / (300 m) = 10⁶ s^-1 = 10⁶ Hz.
5. In this question, we are concerned with how light traveling through some material appears to slow to a velocity of 7.5×10⁷ ms^-1 (I chose 7.5×10⁷ ms^-1 instead of 7.5×10⁸ ms^-1 because the latter is physically unreasonable, being greater than the vacuum speed of light) whilst traveling through the material. The material is, in a way, slowing the light by causing it to bounce off of the atoms located in the materials over and over as it attempts to escape the material. The material is uniform for our intents and purposes. Let v be the speed of light in the material. Observe that 0 < v < c. It is reasonable to think that there should be a relationship between v and c, and there is: they are related by the index of refraction, n. Since the speed of light in a vacuum is constant and also the maximum velocity of light (and of anything in the Universe, it seems for now), we know that 1 < c/v < ∞ (just try different values for v to convince yourself of this if you are unsure). It turns out that n = c/v, so we have for our answer:
6. n = c/v = (3×10⁸ ms^-1) / (7.5×10⁷ ms^-1) = 4.

Answer 2

1) The basic equation for wavelength is equal to velocity divided by frequency. So in this case we have for wavelength (lambda) is equal to 3×10 to the eighth divided by 4× to the eighth, giving us .75 Hz.

2) So given they wavelength of 300 m, you can calculate the frequency by dividing the speed of light by 300 m, giving a frequency equal to 1 MHz or (1×10 to the 6^th Hz).

3) The index of refraction of a material is just the speed of light divided by the velocity of light in the specific material. So here we would divide 3×10 to the eighth by 7.5×10 to the eighth and get an index of refraction equal to 0.4