A light beam travels at 1.94 * 10^8 m/s in quartz. The wavelength of the light in quartz is 355 nm. (b) If this same light travels through air, what is its wavelength there?

Video transcript

Hi everyone. In this practice problem, we're being asked to determine the wavelength of a light weight. Once it emerges into the air, we will have a wavelength of a laser beam in fluoride to be 650 nanometer. The laser beam propagates in fluoride at a speed of two point oh nine times 10 to the power of eight m per second. And we want to find the wavelength of this light once it emerges into the air. So for this particular practice problem, the options given are a 271 nanometer B 427 nanometer C nanometer and D 1398 nanometer. So light travels at its maximum speed C in the vacuum. It is convenient to define the index of reflection of N of a medium to be the ratio of the speed of light in vacuum divided by the speed of light in a particular medium. So N will equals to the speed of light in vacuum divided by the speed of flight in medium or essentially N will equals to C which is the speed of light in vacuum divided by V which is the speed of light in a medium. So in order for us to actually calculate the wavelength of the light, once it emerges into the air, we wanna utilize the other equation where N is actually going to equals to lambda divided by lambda N. Where in this case, the index of refraction of any medium can be expressed as the ratio of LAMBDA which is the wavelength of light in vacuum. And lambda N which is the wavelength of light in the medium whose index of refraction is N. So let's start with the first equation. We want to find what the index of reflection of fluoride is by the information given in our problem statement. So first N will equals to C divided by V and we know that the speed of light is 3.0 times 10 to the power of eight m per second. And the velocity of the weight of the light in the fluoride is given in the problem statement to be two point oh nine times to the power of eight m per second. So then calculating this, this will give us the index of reflection and fluoride to then be equals to 1.44. So now we wanna utilize this 1.44 and input that into the second equation. So we will have uh 1.44 to be equals to Lambda divided by a Lambda N and what we are interested to find is the Lambda here. So in this case, Lambda is then equals to 1. multiplied by Lambda N. And we know what Lambda N is because it is given in our problem statement. So then lambda N is 650 nanometers. So LAMBDA will be equals to 1.44 multiplied by 615 nanometer. And that will give give us a lambda value of 886 nanometer which is going to be the wavelength of the light wave once it emerges into the air. So lambda or wavelength is 886 nanometer or the wavelength of the laser beam in the vacuum is 886 nanometer which is corresponding to option C. So option C will be the answer to this particular practice problem and that'll be it for this video. If you guys still have any sort of other confusion, please make sure to check out our other lesson videos on similar topics and that'll be it for this video. Thank you.