A light wave of frequency \(5 \times 10^{14}\,\text{Hz}\) moves through water which has a refractive index of \(\frac{4}{3}\). Calculate the wavelength in w...

Answer Details

The problem involves finding the wavelength of a light wave in water, given its frequency and the refractive index of water, and assuming that the velocity of light in air is known. First, we need to use the formula v = fλ, where v is the velocity of light, f is the frequency, and λ is the wavelength. We know that the velocity of light in air is 3 x 10^8 m/s, and the frequency of the light wave in water is 5 x 10^14 Hz. We need to find λ. Second, we can use the formula n = c/v, where n is the refractive index, c is the speed of light in a vacuum (which is the same as the speed of light in air), and v is the velocity of light in the medium (in this case, water). We know that the refractive index of water is 4/3. We can solve for v to get v = c/n. Now we can substitute the value of v into the formula v = fλ to get λ = v/f. We have already found v to be c/n, and we know f to be 5 x 10^14 Hz. Substituting these values, we get: λ = (3 x 10^8 m/s)/(4/3)(5 x 10^14 Hz) = 4.5 x 10^-7 m Therefore, the wavelength of the light wave in water is 4.5 x 10^-7 meters. In simple terms, we can say that the problem asks us to find the distance between two consecutive peaks or troughs of a light wave in water, given its frequency and the refractive index of water. To solve the problem, we use formulas that relate the velocity, frequency, wavelength, and refractive index of light in air and water.