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

Question

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 70 degrees. The depth of the lake is 4.3 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?

in progress 0
Maris 2 months 2021-07-19T07:44:46+00:00 1 Answers 5 views 0

Answers ( )

    0
    2021-07-19T07:45:53+00:00

    Answer:

    A) d = 11.8m

    B) d = 4.293 m

    Explanation:

    A) We are told that the angle of incidence;θ_i = 70°.

    Now, if refraction doesn’t occur, the angle of the light continues to be 70° in the water relative to the normal. Thus;

    tan 70° = d/4.3m

    Where d is the distance from point B at which the laser beam would strike the lakebottom.

    So,d = 4.3*tan70

    d = 11.8m

    B) Since the light is moving from air (n1=1.00) to water (n2=1.33), we can use Snell’s law to find the angle of refraction(θ_r)

    So,

    n1*sinθ_i = n2*sinθ_r

    Thus; sinθ_r = (n1*sinθ_i)/n2

    sinθ_r = (1 * sin70)/1.33

    sinθ_r = 0.7065

    θ_r = sin^(-1)0.7065

    θ_r = 44.95°

    Thus; xonsidering refraction, distance from point B at which the laser beam strikes the lake-bottom is calculated from;

    d = 4.3 tan44.95

    d = 4.293 m

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )