Question

Morty is depressed. One night while sitting on the pier, aimlessly shining his laser into the water, he sees a glint from a shiny object. Specifically, he sees this light when he shines the laser 2.4 m in front of him while holding the pointer 1.8 m above the water. If the water is 5.5 m deep, what is the horizontal distance to the shiny object on the ocean floor? (Note: the index of refraction for water is 1.33)

Answer 1

Question #63826, Physics / Optics

Morty is depressed. One night while sitting on the pier, aimlessly shining his laser into the water, he sees a glint from a shiny object. Specifically, he sees this light when he shines the laser 2.4 m in front of him while holding the pointer 1.8 m above the water. If the water is 5.5 m deep, what is the horizontal distance to the shiny object on the ocean floor? (Note: the index of refraction for water is 1.33)

Solution

$$\tan i = \frac {x _ {1}}{y _ {1}} = \frac {2 . 4}{1 . 8} = 1. 3 3; i = 5 3. 1 3 {}^ {\circ}; \sin i = 0. 8$$ $$\sin r = \frac {\sin i}{n} = \frac {0 . 8}{1 . 3 3} = 0. 6 0 2; r = 3 6. 9 8 {}^ {\circ}$$ $$x _ {2} = y _ {2} \tan r = 5. 5 \cdot 0. 7 5 3 = 4. 1 4 \mathrm {m}$$

Horizontal distance from Morty to the shiny object on the ocean floor is $2.4 + 4.14 = 6.54\mathrm{m}$

Answer: 6.54 m.

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