Light of a certain frequency has a wavelength of 526 nm in water. What is the wavelength of this light in benzene?

Video transcript

Hi everyone in this practice problem, we're being asked to determine the wavelength of a light in ethanol. We will have a monochromatic beam with a light of frequency F split into two beams. The first beam has a wavelength of 450 nanometer in a quartz medium with a refractive index of 1.50. The second beam enters an ethanol medium with a refractive index of 1.36. We're being asked to determine the wavelength of this light in ethanol and the options given are a 300 nanometer P, 331 nanometer C 496 nanometer and lastly D 918 nanometer. So the index of refraction of any medium can actually be expressed as the ratio of N to be equals to lambda divided by a lambda of N lambda. Here is the wavelength of light in vacuum. And lambda N is the wavelength of light in the medium whose index of refraction or whose end is N. Therefore, this uh this can be rearranged that we will get the wavelength of light in vacuum to equals to B the refraction index multiplied by the wavelength of that light in the medium of whose index is N. So looking at the two beams, we can actually utilize or equalize the lambda or the wavelength of light in vacuum to be equals to the refractive index of the ethanol multiplied by the wavelength of the light in ethanol equals to the reflective index of the quartz medium multiplied by the wavelength of the quartz medium. So an ethanol multiplied by the lambda of ethanol will then be equals to the end of the quartz multiplied by lambda of the quartz as well. So we can rearrange this equation so that we will get the Lambda for ethanol so that we will get lambda ethanol to be end of quartz multiplied by the lambda of quartz divided by the end of ethanol. Awesome. So we actually know all of this information. So we will just substitute all of this information into lambda ethanol, which will then be equals to N quartz which is 1.50 multiplied by lambda quartz which is 450 nanometer. We will just leave it as nanometer because our answer options are in nanometer as well divided by uh N of ethanol which is 1.36. And that will give us the lambda of ethanol to be 96 nanometer. So the wavelength of this particular light in ethanol is going to be 496 nanometer which will correspond with option C in our answer choices. So option C will be the answer to this particular practice problem and that will be the answer for this video. So if you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics. But other than that, that will be it for this one. Thank you.