A submarine is 2.92 10 2 m horizontally from shore and 1.00 10 2 m beneath the s
ID: 1433352 • Letter: A
Question
A submarine is 2.92 102 m horizontally from shore and 1.00 102 m beneath the surface of the water. A laser beam is sent from the submarine so that the beam strikes the surface of the water 2.14 102 m from the shore. A building stands on the shore, and the laser beam hits a target at the top of the building. The goal is to find the height of the target above sea level. (The index of refraction of water is n = 1.333.)
(a) Draw a diagram of the situation, identifying the two triangles that are important to finding the solution.
(b) Find the angle of incidence of the beam striking the water–air interface. (Give your answer to at least one decimal place.)
____ °
(c) Find the angle of refraction. (Give your answer to at least one decimal place.)
____ °
(d) What angle does the refracted beam make with respect to the horizontal? (Give your answer to at least one decimal place.)
____ °
(e) Find the height of the target above sea level.
____ m
Explanation / Answer
here we can find out it by the help of snell's law.
n1sin(theta1)=n2 sin(theta2)
refractive index of air =1
refractive index of water=1.33
draw a picture it will help you.you can apply tan theta also.
here adjacent side is 100 and opposite side (292-214=78)
hypotinus=sqrt(100^2+78^2)
theta1=adj/hyp
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