Kerala KEAM 5 June 2024 Solved Paper

The refractive index $n$ of a medium is related to the speed of light in vacuum $c$, the speed of light in the medium $v$, and the wavelength of light in the medium $\lambda$ and in vacuum $\lambda_0$ as:$$n = \frac{c}{v} = \frac{\lambda_0}{\lambda}$$Where:- $c$ is the speed of light in vacuum ($3 \times 10^8 \, m/s$),- $\lambda_0$ is the wavelength in vacuum,- $\lambda$ is the wavelength in the medium.From the problem, we are given:- The frequency $f = 4 \times 10^{14} \, Hz$,- The wavelength in the medium $\lambda = 450 \, nm = 450 \times 10^{-9} \, m$.Now, we know that the speed of light in a medium is related to the frequency and wavelength by the equation:$$v = f \lambda$$Substitute the values for frequency and wavelength in the medium:$$v = (4 \times 10^{14}) \times (450 \times 10^{-9}) = 1.8 \times 10^8 \, m/s$$Next, using the equation for refractive index:$$n = \frac{c}{v} = \frac{3 \times 10^8}{1.8 \times 10^8} = 1.67$$Thus, the refractive index of the medium is $\boxed{1.67}$. Quick Tip: The refractive index is the ratio of the speed of light in vacuum to the speed of light in a medium. It can also be determined using the ratio of the wavelength in vacuum to the wavelength in the medium.