Questions: A ray of light enters a pane of glass with an angle of incidence of 37°. The index of refraction of the glass is 1.48. What is the angle of incidence as the ray reaches the glass-air interface on the far side of the pane coming from inside the glass?

Solution Steps

We need to determine the angle of incidence as the ray of light reaches the glass-air interface on the far side of the pane. The initial angle of incidence is given as \(37^\circ\) and the index of refraction of the glass is 1.48.

When the ray of light enters the glass from the air, we use Snell's Law: \[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \] where \(n_1 = 1\) (index of refraction of air), \(\theta_1 = 37^\circ\) (angle of incidence), \(n_2 = 1.48\) (index of refraction of glass), and \(\theta_2\) is the angle of refraction inside the glass.

\[ 1 \cdot \sin 37^\circ = 1.48 \cdot \sin \theta_2 \]

First, calculate \(\sin 37^\circ\): \[ \sin 37^\circ \approx 0.6018 \]

Now, solve for \(\sin \theta_2\): \[ 0.6018 = 1.48 \cdot \sin \theta_2 \] \[ \sin \theta_2 = \frac{0.6018}{1.48} \approx 0.4066 \]

Next, find \(\theta_2\): \[ \theta_2 = \arcsin(0.4066) \approx 24.02^\circ \]

When the ray reaches the glass-air interface on the far side, the angle of incidence inside the glass will be the same as the angle of refraction calculated above, due to the symmetry of the problem.

\[ \theta_2 = 24.02^\circ \]

Final Answer