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# When light undergoes refraction at the surface of separation of two media, what happens to its wavelength?

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## The speed of light in a medium$\left(v\right)$ is a product of the frequency$\left(f\right)$ and wavelength of light$\left(\lambda \right)$.Mathematically it is given as$v=f×\lambda ...\left(1\right)$Now the speed of light depends upon the refractive index$\left(n\right)$ of the medium with the relation $v=\frac{c}{n}...\left(2\right)$, where $c$ is the speed of light.By equating equation $\left(1\right)$and $\left(2\right)$, we get $f×\lambda =\frac{c}{n}\phantom{\rule{0ex}{0ex}}\lambda =\frac{c}{n×f}$The frequency of light does not change when light undergoes refraction, i.e. $f$is constant.Therefore, $\lambda \alpha \frac{1}{n}$For the optically denser medium, the value of $n$is high and for an optically rarer medium $n$is less.Hence, the wavelength decreases when light enters a denser medium, and the wavelength increases when light enters a rarer medium.  Suggest Corrections  2      Similar questions
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