The frequency of a light wave is the same when the light travels in ethyl alcohol or in carbon disulfide.
Find the ratio of the wavelength of the light in ethyl alcohol to that in carbon disulfide.
The refractive indices n of carbon disulfide and ethyl alcohol are n(disulfide) = 1.632 and n(alcohol) = 1.362.
Of course the frequency is the same. There is no box to store extra waves in so the same number of waves have to pass you in a second in each medium.
Therefore:
the wavelength will be proportional to the speed in each medium.
v = c/n
so v in alcohol = c/1.362
and v in disulfide = c/1.632
so v al/vdis = 1.632/1.362
To find the ratio of the wavelength of light in ethyl alcohol to that in carbon disulfide, we can use the formula:
n1λ1 = n2λ2
Where n1 and n2 are the refractive indices of the two mediums, and λ1 and λ2 are the respective wavelengths of light in those mediums.
Given:
n(disulfide) = 1.632
n(alcohol) = 1.362
Let's assume that the wavelength of light in ethyl alcohol is λ(alcohol), and the wavelength of light in carbon disulfide is λ(disulfide).
We can rewrite the formula as:
λ(disulfide) = λ(alcohol) * (n(alcohol)/n(disulfide))
Substituting the given values:
λ(disulfide) = λ(alcohol) * (1.362/1.632)
Simplifying further:
λ(disulfide) = λ(alcohol) * 0.835
Now, the ratio of the wavelength of light in ethyl alcohol to that in carbon disulfide is given by:
λ(alcohol) / λ(disulfide) = 1 / 0.835
Calculating the ratio:
λ(alcohol) / λ(disulfide) ≈ 1.197
Therefore, the ratio of the wavelength of light in ethyl alcohol to that in carbon disulfide is approximately 1.197.
To find the ratio of the wavelength of light in ethyl alcohol to that in carbon disulfide, we can start by using the formula for the speed of light in a medium:
v = c/n,
where v is the speed of light in the medium, c is the speed of light in a vacuum (approximately 3 x 10^8 meters per second), and n is the refractive index of the medium.
Since we have the refractive indices for ethyl alcohol and carbon disulfide, we can find their respective speeds of light:
v(ethanol) = c / n(ethanol),
v(disulfide) = c / n(disulfide).
We are given that the frequencies of the light waves are the same in both media. Since the speed of light is determined by the product of its frequency and wavelength (v = λf), this means that the speed of light is constant, regardless of the medium it is traveling through.
Therefore, we can equate the speeds of light in ethyl alcohol and carbon disulfide:
v(ethanol) = v(disulfide).
Substituting the expressions for the speeds of light, we get:
c / n(ethanol) = c / n(disulfide).
Now, rearranging for the ratio of the wavelengths:
λ(ethanol) / λ(disulfide) = n(disulfide) / n(ethanol).
Plugging in the values for the refractive indices:
λ(ethanol) / λ(disulfide) = 1.632 / 1.362.
Calculating this ratio, we find:
λ(ethanol) / λ(disulfide) = 1.198.
Therefore, the ratio of the wavelength of light in ethyl alcohol to that in carbon disulfide is approximately 1.198.