Light is incident on a glass block from air. If the angle between the reflected ray and the refracted ray is 90 degrees, find the angle of incidence.

The refractive index of air (medium 1) is $n_1 = 1$.
The refractive index of the glass block (medium 2) is $n_2 = \frac{3}{2} = 1.5$.
The angle between the reflected ray and the refracted ray is $90^\circ$. Let $\theta_i$ be the angle of incidence, $\theta_r$ be the angle of reflection, and $\theta_t$ be the angle of refraction.
According to the law of reflection, the angle of incidence equals the angle of reflection: $\theta_i = \theta_r$.
The given condition is $\theta_r + \theta_t = 90^\circ$.
Substituting $\theta_r = \theta_i$, we get $\theta_i + \theta_t = 90^\circ$.
From this, we can express the angle of refraction as $\theta_t = 90^\circ - \theta_i$.

Snell's Law states:

Substitute the known values and the expression for $\theta_t$:

We know that $\sin(90^\circ - x) = \cos(x)$. So,

Divide both sides by $\cos(\theta_i)$:

To find $\theta_i$, take the inverse tangent of 1.5:

The angle of incidence is approximately $\text{56.31}^\circ$.