Question

A submarine is 2.99 ✕ 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. (Submit a file with a maximum size of 1 MB.) image.jpeg This answer has not been graded yet.(b) Find the angle of incidence of the beam striking the water–air interface. (Give your answer to at least one decimal place.) Correct: Your answer is correct. °c) Find the angle of refraction. (Give your answer to at least one decimal place.) Incorrect: Your answer is incorrect. °(d) What angle does the refracted beam make with respect to the horizontal? (Give your answer to at least one decimal place.) Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations.°(e) Find the height of the target above sea level. Correct: Your answer is correct. m

Answer 1

Answer:

a The diagram of the situation is shown on the first uploaded image

b the angle of  incidence  beam striking the water is $\theta = 49.63^o$

c  the angle of  refraction  beam striking the water is $r = 59.7^o$

d The angle the refracted beam make with respect to the horizontal is $= 30.3^o$

e The height of the target above sea level is  

                   $h= 125.05m$

Explanation:

From the diagram we see that the angle of the beam striking the water is

                   $tan \theta = \frac{100}{85}$

                        $\theta = tan^{-1}(\frac{100}{85} )$

                           $= 49.63^o$

According to Snell's law

                   $\mu_{water} *sin(i) = \mu_{air} *sin(r)$

Where $\mu_{water }$ is the refractive index of water =  1.333

           $i$ is the angle of incidence

          $\mu_{air}$ is the refractive index of air  = 1

            r is the angle of refraction

 Substituting values accordingly

          $1.33 * sin (40.37) = 1 * sin(r)$

    Making r the subject of the formula

                       $r = sin^{-1}(\frac{1.333 *sin(40.37)}{1})$

                          $= 59.7^o$

looking at the diagram we can see that to  obtain the angle the refraction beam makes with the horizontal   by subtracting the angle refraction from 90°

                 i.e  $90 -59.7 = 30.3$ °

From the diagram we see that the height  target above sea level can be obtained by this relation

                   $tan \theta = \frac{h}{214}$

Where h is the is the height

                   $tan(30.3) = \frac{h}{214}$

                         $h = 214 * tan (30.3)$

                            $=125.05m$