In the figure,a laser positioned on a ship is used to communicate with a small two-man research submarine resting on the bottom of a lake.The laser is positioned 12 m above the surface of the water,and it strikes the water 20 m from the side of the ship.The water is 76 m deep and has an index of refraction of 1.33.How far horizontally is the submarine from the side of the ship?
A)84.1 m
B)64.1 m
C)104 m
D)74.1 m
E)94.1 m

Question

In the figure,a laser positioned on a ship is used to communicate with a small two-man research submarine resting on the bottom of a lake.The laser is positioned 12 m above the surface of the water,and it strikes the water 20 m from the side of the ship.The water is 76 m deep and has an index of refraction of 1.33.How far horizontally is the submarine from the side of the ship? 
A)84.1 m 
B)64.1 m 
C)104 m 
D)74.1 m 
E)94.1 m

Quizplus · Accepted Answer

First, we need to calculate the angle of incidence of the laser beam at the surface of the water. Using trigonometry:

$\sin 	heta_1 = \frac{12}{20}$
$	heta_1 = \sin^{-1} \frac{12}{20} \approx 36.9^\circ$

Next, we need to calculate the angle of refraction of the laser beam at the water-air boundary. Using Snell's Law:

$n_1 \sin 	heta_1 = n_2 \sin 	heta_2$
$(1.00) \sin 36.9^\circ = (1.33) \sin 	heta_2$
$	heta_2 \approx 28.3^\circ$

Now, we can use the angle of refraction and the depth of the water to calculate the distance from the point where the laser beam enters the water to the point where it reaches the submarine. Using trigonometry:

$\sin 	heta_2 = \frac{d}{76}$
$d = 76 \sin 	heta_2 \approx 34.1 	ext{ m}$

Finally, we can use the angle of incidence and the distance from the ship to the point where the laser beam enters the water to calculate the horizontal distance from the ship to the submarine. Using trigonometry:

$\cos 	heta_1 = \frac{h}{d + x}$
$x = (d + h) \cos 	heta_1 - d \approx 84.1 	ext{ m}$

Therefore, the answer is (A) 84.1 m.

In the Figure,a Laser Positioned on a Ship Is Used