A beam of light has a wavelength of 650 nm in vacuum. (b) What is the wavelength of these waves in the liquid?

Video transcript

Hi everyone in this practice problem, we are being asked to determine the wavelength of the blue light in methanol. We will have a blue light ray propagating in a transparent methanol liquid with a refractive index of 1.33. The wavelength of the blue light in a vacuum is 455 nanometer. And we're being asked to determine the wavelength of the blue light in methanol. And the options given are a 100 and 55 nanometer B 342 nanometer C 605 nanometer and lastly D 978 nanometer. So for us to be able to solve this problem, we have to recall that the index of refraction of any medium can be expressed as the ratio of N to E equals to lambda divided by a lambda N where in this case, lambda is the wavelength of light in vacuum. And lambda N is the wavelength of light in the medium whose index of refraction is N. In this particular practice problem, we are talking about methanol and the end of methanol is given to be 1.33. So in this case, 1.33 or the end of methanol will be equals to the LAMBDA or the wavelength of light in vacuum divided by the wavelength of light in methanol. So LAMBDA and we were given that the blue light in vacuum is for is going to have the wavelength of 455 nanometer. So we wanna uh rearrange this equation to get the equation for the wavelength of the blue light in methanol or Lambda N, which will then be the lambda divided by a 1.33. So let's substitute the Lambda here which is going to be 455 nanometer which is the uh wavelength of the blue light in methanol divided by 1.33 to get the lambda and or the wavelength of the blue light in Methanol to be 342 nanometer. So the wavelength of the glue and methanol is 342 nanometer which will be corresponding to option B in our answer choices, which will be the answer to this practice problem. So that will be it for this video. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that will be it for this one. Thank you.