If $\hat{i}$ denotes a unit vector along an incident ray, $\hat{r}$ is unit vector along the refracted ray into a medium of refractive index ' $^{'} μ^{'}$ and $\hat{n}$ a unit vector normal to the boundary of the media directed towards the incident medium, then the law of refraction can be written as

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$\hat{i} \times \hat{n} = μ (\hat{n} \times \hat{r})$

$\hat{i} \times \hat{n} = μ (\hat{r} \times \hat{n})$

$μ (\hat{i} \times n) = \hat{r} \times \hat{n}$

$\hat{i} . \hat{n} = μ (\hat{r . n})$

answer is C.

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Detailed Solution

$\begin{array}{l} \bar{A} \times \bar{B} = | \bar{A} | | \bar{B} | \sin θ \\ \hat{i} \times \hat{n} = | \hat{i} | | | \hat{r} ∣ \sin i = 1 \times 1 \times \sin i = \sin i \\ \hat{r} \times \hat{n} = | \hat{r} | | \hat{n} | \sin r = 1 \times 1 \times \sin r = \sin r \\ μ = \frac{\sin i}{\sin r} = \frac{\hat{i} \times \hat{n}}{\hat{r} \times \hat{n}} ∴ \hat{i} \times \hat{n} = μ (\hat{r} \times \hat{n}) \end{array}$