Given that $n_1 = 1$ and $n_2 = 1.55$, we can rearrange the equation to solve for $\theta_1$:
$$\sin(\theta_1) = \frac{n_2}{n_1} \sin(\theta_2)$$
$$\sin(\theta_1) = \frac{1.55}{1} \sin(90^\circ)$$
$$\sin(\theta_1) = 1.55$$
$$\theta_1 = \sin^{-1}(1.55)$$
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