## Step 1: Find the Angle of Refraction (Snell's Law) **Given:** - Angle of incidence, \(\theta_1 = 30.^\circ\) - Refractive index of glass, \(n = 1.5\) - Refractive index of air, \(n_1 \approx 1.\) **Snell's Law:** \[ n_1 \sin\theta_1 = n_2 \sin\theta_2 \] \[ 1. \times \sin 30^\circ = 1.5 \times \sin\theta_2 \] \[ .5 = 1.5 \sin\theta_2 \] \[ \sin\theta_2 = \frac{.5}{1.5} = .333 \] \[ \theta_2 = \arcsin(.333) \approx 19.5^\circ \] **Explanation:** We used Snell's Law to find the angle of refraction (\(\theta_2\)) inside the glass block. --- ## Step 2: Find the Lateral Shift \(d\) **Given:** - Thickness of glass block, \(t = 2.00\, \text{cm}\) - \(\theta_1 = 30^\circ\), \(\theta_2 \approx 19.5^\circ\) **Lateral shift formula:** \[ d = t \frac{\sin(\theta_1 - \theta_2)}{\cos\theta_2} \] \[ d = 2.00\,\text{cm} \times \frac{\sin(30^\circ - 19.5^\circ)}{\cos(19.5^\circ)} \] \[ d = 2.00\,\text{cm} \times \frac{\sin(10.5^\circ)}{\cos(19.5^\circ)} \] \[ \sin(10.5^\circ) \approx .182, \quad \cos(19.5^\circ) \approx .942 \] \[ d = 2.00\,\text{cm} \times \frac{.182}{.942} \approx 2.00\,\text{cm} \times .193 \approx .386\,\text{cm} \] **Explanation:** We used the lateral shift formula for a slab to find the value of \(d\). --- ## **Final Answer** \[ \boxed{d \approx .39\,\text{cm}} \]