JAMB Physics 2023 Paper

Concept:When light enters a different medium, its frequency remains unchanged; only its speed and wavelength change.Explanation:The refractive index is given as $n = 1.458$.The speed of light in fused quartz is $v = \frac{c}{n}$.So $v = \frac{3.00 \times 10^8}{1.458} \approx 2.06 \times 10^8\ \mathrm{m\,s}^{-1}$.The wavelength in the medium is $\lambda = \frac{\lambda_0}{n}$.Therefore $\lambda = \frac{589 \times 10^{-9}}{1.458} \approx 4.04 \times 10^{-7}\ \mathrm{m}$.Using the wave equation $v = f\lambda$, the frequency is $f = \frac{v}{\lambda}$.Substitute the values: $f = \frac{2.06 \times 10^8}{4.04 \times 10^{-7}}$.This gives $f \approx 5.09 \times 10^{14}\ \mathrm{Hz}$.The same result can be obtained directly using $f = \frac{c}{\lambda_0}$, confirming that the frequency does not depend on the medium.Answer:B. $5.09 \times 10^{14}\ \mathrm{Hz}$