Answer
The refracted light is white dominated by the red end.
Work Step by Step
We can use Snell's law to find $\theta_r$, the angle of refraction of red light:
$n_2~sin~\theta_r = n_1~sin~\theta_1$
$sin~\theta_r = \frac{n_1~sin~\theta_1}{n_2}$
$\theta_r = sin^{-1}~(\frac{n_1~sin~\theta_1}{n_2})$
$\theta_r = sin^{-1}~(\frac{1.456~sin~43.10^{\circ}}{1.00})$
$\theta_r = 84.2^{\circ}$
We can use Snell's law to find $\theta_b$, the angle of refraction of blue light:
$n_2~sin~\theta_b = n_1~sin~\theta_1$
$sin~\theta_b = \frac{n_1~sin~\theta_1}{n_2}$
$sin~\theta_b = \frac{1.470~sin~43.10^{\circ}}{1.00}$
$sin~\theta_b = 1.004$
Since there is no angle $\theta_b$ such that $sin~\theta_b = 1.004$, the blue end of the spectrum is not refracted.
The refracted light is white dominated by the red end.
