# Solution Let's solve each part step by step. --- ## (a) Horizontal Distance \( d \) from the Man to the Edge (Empty Ditch) ### **Step 1: Draw and Analyze the Geometry** - The man’s eye is at height \( h = 1.68 \) m above the ground. - The ditch is \( 2.80 \) m deep. - The line of sight just grazes the far edge at the bottom, making an angle \( \theta = 32.7^\circ \) below the horizontal. - The vertical drop from his eye to the bottom of the ditch is: \[ \text{Vertical drop} = 2.80\,\text{m} - 1.68\,\text{m} = 1.12\,\text{m} \] - The horizontal distance \( d \) is adjacent to \( \theta \) in the right triangle formed. ### **Step 2: Use Trigonometry to Find \( d \)** \[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{1.12}{d} \] \[ d = \frac{1.12}{\tan 32.7^\circ} \] \[ \tan 32.7^\circ \approx .6427 \] \[ d = \frac{1.12}{.6427} \approx 1.743\,\text{m} \] **Explanation:** We used the tangent of the given angle, with the vertical drop as the opposite side and \( d \) as the adjacent side. --- ### **Final Answer (a):** \[ \boxed{1.74\,\text{m}} \] --- ## (b) Maximum Distance \( x \) with Water in the Ditch ### **Step 1: Consider Refraction at the Water Surface** - Now, light coming from the boundary at the bottom passes from water (\( n = 1.33 \)) to air (\( n = 1.00 \)) and reaches the man's eye at angle \( \theta = 32.7^\circ \). - Let \( \theta_2 = 32.7^\circ \) (in air, to the man's eye). - Use Snell's Law to find the angle in water (\( \theta_1 \)): \[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \] \[ 1.33 \sin \theta_1 = 1.00 \sin 32.7^\circ \] \[ \sin \theta_1 = \frac{\sin 32.7^\circ}{1.33} \approx \frac{.5403}{1.33} \approx .4065 \] \[ \theta_1 = \arcsin(.4065) \approx 23.97^\circ \] **Explanation:** We used Snell's law to find the angle the light must make in water to emerge at \( 32.7^\circ \) in air. --- ### **Step 2: Find the Maximum Distance \( x \) Using Trigonometry** - In water, the ray travels from the boundary (bottom/far edge) to the water surface (near edge). - The horizontal distance in water (\( d' \)) is: \[ d' = \frac{2.80}{\tan \theta_1} \] \[ \tan 23.97^\circ \approx .4442 \] \[ d' = \frac{2.80}{.4442} \approx 6.304\,\text{m} \] - In air, the vertical height from the ground to the eye is \( h = 1.68 \) m, and the horizontal distance in air (\( d'' \)) is: \[ d'' = \frac{1.68}{\tan 32.7^\circ} = \frac{1.68}{.6427} \approx 2.615\,\text{m} \] - The total maximum distance from the man to the edge: \[ x = d' + d'' = 6.304 + 2.615 = 8.919\,\text{m} \] **Explanation:** We calculated the two segments of light path (underwater and in air) and added them for the total maximum distance. --- ### **Final Answer (b):** \[ \boxed{8.92\,\text{m}} \]