A spotlight on a boat is y = 2.4 m above the water, and the light strikes the wa

Question

A spotlight on a boat is y = 2.4 m above the water, and the light strikes the water at a point that is x = 9.2 m horizontally displaced from the spotlight (see the drawing). The depth of the water is 4.0 m. Determine the distance d, which locates the point where the light strikes the bottom.

Explanation / Answer

OK, ........................... Spot Water ________/____Boat The angle between water and / light going in is ARCTAN (2.8/8.8) which is 17.65012deg So, from the normal for the incident ray being 72.3499deg, use Snell's law (REMEMBER SNELL'S LAW REQUIRES USE OF "NORMAL" RAYS) -> sin I / sin R = refractive index, n n for air/water = 4/3 So, (sin (72.34....)) / (4/3) is sin R sin R = 0.71469... and R = 45.618...deg. The inner angle of a triangle under the water (to the bottom of the water, 4m lower) is then 44.382...deg Calculating using trig again, the "adjacent" is then 4m / (tan 44.382..) = 4.087m So the total distance is from the boat (8.8) + the horizontal light travel under the water (4.087) =12.887m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.