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(A) The wavelength of light in a medium is given by $\lambda_n = \frac{\lambda_0}{n}$,where $\lambda_0$ is the wavelength in vacuum and $n$ is the ref

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$n_3 > n_1 > n_2$

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(A) The wavelength of light in a medium is given by $\lambda_n = \frac{\lambda_0}{n}$,where $\lambda_0$ is the wavelength in vacuum and $n$ is the refractive index of the medium.From the relation $\lambda_n \propto \frac{1}{n}$,a smaller wavelength corresponds to a higher refractive index.By observing the provided image,we can compare the wavelengths in the three regions:In region $n_2$,the wave has the longest wavelength ($\lambda_2$ is largest).In region $n_1$,the wavelength $(\lambda_1)$ is shorter than $\lambda_2$.In region $n_3$,the wavelength $(\lambda_3)$ is the shortest.Therefore,the order of wavelengths is $\lambda_2 > \lambda_1 > \lambda_3$.Since the refractive index is inversely proportional to the wavelength,the order of refractive indices is $n_2  n_1 > n_2$.

$A$ light wave travels through three transparent materials of equal thickness. Rank in order,from the highest to lowest,the indices of refraction $n_1, n_2$ and $n_3$.

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