A submarine is 3.00 x 10^2 m horizontally from shore and 120.0 m beneath the sur

Question

A submarine is 3.00 x 10^2 m horizontally from shore and 120.0 m beneath the surface of the water. A laser beam is sent from the submarine so that the beam strikes 2.10 x 10^2 m from the buildingstanding on the shoreand the laser beam hits the target on the top of the building. Use n=1.333 for the refractive index of water.A)Calculate the angle of incidence ?1 on the water/air interface.

B)Calculate the angle of refraction ?2.
C)Find angle ? with the horizontal.
D)Find the height h of the building

Explanation / Answer

know that the submarine is 300 m from the building on the shore

and that the beam from the submarine strikes 210 from the building standing on the shore

part a

so 300-210 = d = 90

use trig to get theta1 and knowing that sub is 120m below surface of water

you got tan (theta1) = opp/ hyp = 90/120 = 0.75

arctan 0.75 = ( theta 1)

36.869 degrees is = (theta1) which is your angle of incidence

PART b

next use Snells law to get angle of refraction (theta 2)

n1sin(theta 1) = n2 sin (theta 2)

isolating for sin (theta 2) you get

n1sin (theta 1) / n2 = sin ( theta 2) your n1 is water = 1.333 and n2 is air = 1.0

1.333sin(36.869) / 1 = sin (theta 2)

0.79978 = sin (theta 2)

arcsin(0.79978) = (theta 2) which s your angle of refraction

53.1 degrees = (theta 2) which s your angle of refraction

pART c

Phi = 90 - ( theta 2)

phi = 90 - 53.1

phi = 36.89 degrees

PART D

lastly solving for height

tan( phi) = h / 210m

Tan (phi) X 210 m = H

tan (36.89) x 210 = H

157.61 m = H

hope that helped

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