A narrow beam of light from a laser travels through air (n = 1.00) and strikes the surface of the water (n = 1.33) in a lake at point A. The
angle of incidence is 52 degrees. The depth of the lake is 4.5 m. On the flat lake-bottom is point B, directly below point A. (a) If refraction did not occur, how far away from point B would the laser beam strike the lakebottom? (b) Considering refraction, how far away from point B would the laser beam strike the lake-bottom?
(a)
Let x be the distance from point B to the point where the laser beam would strike the lake-bottom if refraction did not occur.
Using the formula for Snell's Law: n1sinθ1 = n2sinθ2
Where n1 = 1.00 (air), n2 = 1.33 (water), and θ1 = 52 degrees
Calculating θ2 using Snell's Law:
1.00 * sin(52 degrees) = 1.33 * sin(θ2)
sin(θ2) = 0.7744
θ2 ≈ 51.2 degrees
Now, considering the triangle formed by the laser beam, the depth of the lake (4.5 m), and the point where the laser beam would strike the lake-bottom:
tan(θ2) = 4.5/x
x = 4.5 / tan(51.2 degrees)
x ≈ 3.45 meters
Therefore, if refraction did not occur, the laser beam would strike the lake-bottom approximately 3.45 meters away from point B.
(b)
Considering refraction, we can use the law of refraction to calculate the distance at which the laser beam would strike the lake-bottom.
Snell's Law: n1sinθ1 = n2sinθ2
Where n1 = 1.00 (air), n2 = 1.33 (water), and θ1 = 52 degrees
Calculating θ2 using Snell's Law:
1.00 * sin(52 degrees) = 1.33 * sin(θ2)
sin(θ2) = 0.7744
θ2 ≈ 51.2 degrees
Using the same triangle as in part (a) with the refracted angle of 51.2 degrees:
tan(θ2) = 4.5/x
x = 4.5 / tan(51.2 degrees)
x ≈ 3.45 meters
Therefore, taking into account refraction, the laser beam would still strike the lake-bottom at approximately 3.45 meters away from point B.