91Ó°ÊÓ

To find the speed of light in water, we can use the formula: $$n = \frac{c}{v}$$ Where \(n\) is the index of refraction, \(c\) is the speed of light in vacuum, and \(v\) is the speed of light in the medium (in this case, water). Rearrange and solve for the speed of light in water (\(v\)): $$v = \frac{c}{n}$$ Now substitute the values, \(n=1.33\) and \(c=3.0\times10^8\)m/s: $$v = \frac{3.0\times10^8\,\text{m/s}}{1.33}$$ Calculate the speed of light in water: $$v \approx 2.26\times10^8\,\text{m/s}$$

To find the wavelength of light in water, we will use the equation: $$v = \lambda'f$$ where \(v=2.26\times10^8\,\text{m/s}\) is the speed of light in water, \(\lambda'\) is the wavelength in water, and \(f\) is the frequency. Since the frequency remains constant when light enters a new medium, we can set up the equation for both vacuum and water: $$c = \lambda f$$ where \(c=3.0\times10^8\,\text{m/s}\) is the speed of light in vacuum and \(\lambda = 515\,\text{nm}\) is the vacuum wavelength. Solve for the frequency, \(f\) from the vacuum equation: $$f = \frac{c}{\lambda}$$ Substitute the values, \(c=3.0\times10^8\,\text{m/s}\) and \(\lambda=515\,\text{nm} = 515\times10^{-9}\,\text{m}:\) $$f \approx 5.83\times10^{14}\,\text{Hz}$$ Now use the equation for light in water and substitute the values, \(v=2.26\times10^8\,\text{m/s}\) and \(f=5.83\times10^{14}\,\text{Hz}\): $$2.26\times10^8\,\text{m/s} = \lambda' \times 5.83\times10^{14}\,\text{Hz}$$ Solve for the wavelength in water \(\lambda'\): $$\lambda' \approx 3.87\times10^{-7}\,\text{m} = 387\,\text{nm}$$ The wavelength of the light in water with vacuum wavelength of \(515\,\text{nm}\) is \(387\,\text{nm}.\)