A monochromatic light of frequency $5 \times 10^{14} \mathrm{~Hz}$ travelling through air, is incident on a medium of refractive index ' 2 '. Wavelength of the refracted light will be :

<p>To find the wavelength of the refracted light in a medium with a refractive index of 2, we can use the relationship between frequency, wavelength, and the speed of light. The frequency ($f$) of the light remains the same when it enters the medium.</p> <p>The formula relating frequency, wavelength, and velocity is:</p> <p>$ v = f \lambda $</p> <p>Where:</p> <p><p>$ v $ is the speed of light in the medium,</p></p> <p><p>$ f $ is the frequency of the light,</p></p> <p><p>$ \lambda $ is the wavelength of the light.</p></p> <p>The wavelength of light in the medium ($\lambda_{\text{medium}}$) can be found using the refractive index ($\mu$) and the wavelength in a vacuum ($\lambda_{\text{vacuum}}$) as follows:</p> <p>$ \lambda_{\text{medium}} = \frac{\lambda_{\text{vacuum}}}{\mu} $</p> <p>Given:</p> <p><p>Frequency of the light, $ f = 5 \times 10^{14} \, \text{Hz} $</p></p> <p><p>Speed of light in a vacuum, $ v_{\text{vacuum}} = 3 \times 10^{8} \, \text{m/s} $</p></p> <p><p>Refractive index of the medium, $ \mu = 2 $</p></p> <p>First, calculate $\lambda_{\text{vacuum}}$:</p> <p>$ \lambda_{\text{vacuum}} = \frac{v_{\text{vacuum}}}{f} = \frac{3 \times 10^{8}}{5 \times 10^{14}} $</p> <p>Now calculate $\lambda_{\text{medium}}$:</p> <p>$ \lambda_{\text{medium}} = \frac{\lambda_{\text{vacuum}}}{2} = \frac{3 \times 10^{8}}{2 \times 5 \times 10^{14}} $</p> <p>$ \lambda_{\text{medium}} = 0.3 \times 10^{-6} \, \text{m} $</p> <p>Converting to nanometers:</p> <p>$ 0.3 \times 10^{-6} \, \text{m} = 300 \, \text{nm} $</p> <p>Therefore, the wavelength of the refracted light in the medium is <strong>300 nm</strong>.</p>