91Ó°ÊÓ

Snell's Law states that the ratio of the sines of the angle of incidence (\(\theta_1\)) and the angle of refraction (\(\theta_2\)) in different media is equal to the inverse ratio of their refractive indices (\(n_1\) and \(n_2\)). From the problem, we know that \(\theta_1 = 60.0^{\circ}\), and for air \(n_1 = 1\). The refractive index of dense flint glass (taken from a reference table) is \(n_2 = 1.66\). Applying Snell's Law: $$\frac{\sin\theta_1}{\sin\theta_2} = \frac{n_2}{n_1}$$ We can then solve for \(\theta_2\): $$\sin\theta_2 = \frac{\sin\theta_1 \cdot n_1}{n_2} = \frac{\sin60.0^{\circ} \cdot 1}{1.66}$$ $$\theta_2 = \arcsin\left(\frac{\sin60.0^{\circ}}{1.66}\right) \approx 32.2^{\circ}$$ The angle of refraction in the glass is \(\approx 32.2^{\circ}\).

We know the refracted angle in the glass, and we can use this information to find the vertical and horizontal displacements of the ray within the glass medium. The total vertical displacement (\(D_v\)) is equal to the thickness of the glass (\(T_g = 5.00 \,\text{mm}\)), and the horizontal displacement (\(D_h\)) can be found using trigonometry. In this case, the trigonometric function \(\tan\) is used: $$D_h = T_g \cdot \tan\theta_2 = 5.00 \,\text{mm} \cdot \tan(32.2^{\circ}) \approx 3.33 \,\text{mm}$$ The horizontal displacement within the glass is \(\approx 3.33 \,\text{mm}\).

The light ray enters and exits the parallel glass surfaces, so the angles of incidence and emergence are the same (\(60.0^{\circ}\) for both). This implies that each segment of the light ray's horizontal path contributes equally to the total horizontal displacement. There are two segments of light ray before and after the glass, so we double the horizontal displacement within the glass to find the total displacement of the light ray: $$D_t = 2 \cdot D_h = 2 \cdot 3.33 \,\text{mm} \approx 6.66 \,\text{mm}$$ The total horizontal displacement of the light ray is \(\approx 6.66 \,\text{mm}\).