Question

Monochromatic light of frequency $5.0 \times 10^{14 \, \text{Hz}$ passes from air into a medium of refractive index $1.5$. Find the wavelength of the light (i) reflected, and (ii) refracted at the interface of the two media.}

Hint:

The wavelength of reflected light does not change as it remains in the same medium. The wavelength of refracted light changes and is inversely proportional to the refractive index of the medium.

Answer 1

The wavelength $\lambda$ of light is related to its speed and frequency by: \[ \lambda = \frac{v}{f}. \] For light in air: \[ v_{\text{air}} = c = 3.0 \times 10^8 \, \text{m/s}. \] The wavelength in air is: \[ \lambda_{\text{air}} = \frac{v_{\text{air}}}{f} = \frac{3.0 \times 10^8}{5.0 \times 10^{14}} = 600 \, \text{nm}. \] % Option (i) The wavelength of the reflected light remains the same as in air: \[ \boxed{\lambda_{\text{reflected}} = 600 \, \text{nm}}. \] % Option (ii) For refracted light in the medium: \[ v_{\text{medium}} = \frac{v_{\text{air}}}{n} = \frac{3.0 \times 10^8}{1.5} = 2.0 \times 10^8 \, \text{m/s}. \] The wavelength in the medium is: \[ \lambda_{\text{refracted}} = \frac{v_{\text{medium}}}{f} = \frac{2.0 \times 10^8}{5.0 \times 10^{14}} = 400 \, \text{nm}. \] Thus: \[ \boxed{\lambda_{\text{refracted}} = 400 \, \text{nm}}. \]