WAEC SSCE - Physics - 1994 (Objective)

Question 1 Report

A light wave travels from air into a medium of refractive index 1.54. If the wavelength of the light in air is \(5.9\times10^{-7}\)m, calculate its wavelength in the medium

\(2.5\times10^{-7}\)m

\(3.8\times10^{-7}\)m

\(4.6\times10^{-7}\)m

\(5.2\times10^{-7}\)m

\(6.1\times10^{-7}\)m

Answer Details

The relationship between the wavelength of light in two different media and their refractive indices is given by the formula: \begin{equation*} \frac{\lambda_1}{\lambda_2} = \frac{n_2}{n_1} \end{equation*} Where: \begin{itemize} \item \(\lambda_1\) is the wavelength of light in the first medium \item \(\lambda_2\) is the wavelength of light in the second medium \item \(n_1\) is the refractive index of the first medium \item \(n_2\) is the refractive index of the second medium \end{itemize} In this case, the light wave travels from air (first medium) into a medium of refractive index 1.54 (second medium). Therefore, we can use the above formula to find the wavelength of the light in the second medium: \begin{align*} \frac{\lambda_1}{\lambda_2} &= \frac{n_2}{n_1} \\ \lambda_2 &= \frac{n_1}{n_2} \lambda_1 \\ &= \frac{1}{1.54} \times 5.9 \times 10^{-7} \mathrm{m} \\ &\approx 3.84 \times 10^{-7} \mathrm{m} \end{align*} Therefore, the wavelength of the light in the medium is approximately \(3.84\times10^{-7}\)m. Answer: \boxed{2}. \(3.8\times10^{-7}\)m.

Read lesson note on Refraction Of Light Through Plane And Curved Surfaces (JAMB)






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