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Chapter 32: Problem 44

A light ray of wavelength 700 . \(\mathrm{nm}\) traveling in air \(\left(n_{1}=1.00\right)\) is incident on a boundary with a liquid \(\left(n_{2}=1.63\right) .\) a) What is the frequency of the refracted ray? b) What is the speed of the refracted ray? c) What is the wavelength of the refracted ray?

Short Answer

Expert verified
Question: A light ray with an initial wavelength of 700 nm travels from air (refractive index 1.00) into a liquid (refractive index 1.63). Calculate the frequency, speed, and wavelength of the refracted ray in the liquid. Answer: The refracted ray has a frequency of \(4.286\times10^{14}\, \mathrm{Hz}\), a speed of \(1.840\times10^8\, \mathrm{m/s}\) in the liquid, and a wavelength of \(430\, \mathrm{nm}\) in the liquid.

Step by step solution

01

Calculate the initial speed and frequency of the light ray in air

Recalling that the refractive index is given by \(n = \frac{c}{v}\), where \(v\) is the speed of light in the medium. We can solve for the speed of light in air as: $$ v_{1} = \frac{c}{n_{1}} $$ Then, we can calculate the frequency of the light ray using the relationship \(v = f \lambda\), where \(f\) is the frequency, and \(\lambda\) is the wavelength: $$ f = \frac{v_{1}}{\lambda_{1}} $$
02

Calculate the speed of the refracted ray in the liquid

Using the refractive index \(n_{2}\), we find the speed of the refracted light ray in the liquid as: $$ v_{2} = \frac{c}{n_{2}} $$
03

Calculate the wavelength of the refracted ray in the liquid

Since the frequency is constant when light travels from one medium to another, we have: $$ f_{1} = f_{2} $$ And by using the relationship \(v = f \lambda\), we can calculate the wavelength of the refracted ray in the liquid as: $$ \lambda_{2} = \frac{v_{2}}{f_{2}} = \frac{v_{2}}{f_{1}} $$ Now let's plug in the given values and solve for the required quantities:
04

Plug in the given values and solve

From steps 1, 2, and 3, we have: $$ v_{1} = \frac{3.00\times10^8\, \mathrm{m/s}}{1.00} = 3.00\times10^8\, \mathrm{m/s} $$ $$ f = \frac{3.000\times10^8\, \mathrm{m/s}}{700\times10^{-9}\, \mathrm{m}} = 4.286\times10^{14}\, \mathrm{Hz} $$ $$ v_{2} = \frac{3.00\times10^8\, \mathrm{m/s}}{1.63} = 1.840\times10^8\, \mathrm{m/s} $$ $$ \lambda_{2} = \frac{1.840\times10^8\, \mathrm{m/s}}{4.286\times10^{14}\, \mathrm{Hz}} = 430\times10^{-9}\, \mathrm{m} $$ So, the refracted ray has: a) a frequency of \(4.286\times10^{14}\, \mathrm{Hz}\), b) a speed of \(1.840\times10^8\, \mathrm{m/s}\) in the liquid, c) a wavelength of \(430\, \mathrm{nm}\) in the liquid.

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