Joseph H. answered • 09/05/21
Physics graduate student with experience teaching physics and maths
These will both take some application of Snell's Law (n1sinθ1=n2sinθ2, note the angles are measured from the normal of the boundary between media 1 and 2) and some triangles.
1a.) For this part, you see the mast making an angle of 33º, and that sight line will make a vertical angle with the one needed for the water side of a Snell's Law situation. The water index of refraction is given, and that of air is about 1, so we have the angle on the other side. Now all we need is one leg of the triangle that includes this angle and the mast height m. We can find this using the distance b from the boat to the point light rays from the top of the mast are hitting the water. This will mean finding how close you are to that point by finding the other leg of the triangle that includes your depth h and subtracting this from the distance to the boat d. This will mean by construction of right triangles that b = d - h tanθ1. Finally, we can construct another triangle with θ2, b, and m, and observe that m = b / tanθ2.
1b.) Here we will again apply Snell's Law. Physically speaking, the highest angle a light ray can make with the normal and "hit" the water's surface is 90º. Based on Snell's Law, these will then make the greatest angle with the normal after refraction as well and so will travel the furthest underwater, thus marking the outer edge of the circle of sky. This consideration makes the right hand side of Snell's law to be just 1, and the equation can then be solved for the angle θsky = arcsin(1/n1). Taking this angle, we construct one more triangle which has legs corresponding to your depth and to the radius r of the circle of sky so that s = h tanθsky.
My numerical results are 3.86m and 6.84m for a and b respectively. Hope this helps!!
