MasteringPhysics: HW 11 Ch. 33 Google Chrome https:// session.masteringphysics.c

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MasteringPhysics: HW 11 Ch. 33 Google Chrome https:// session.masteringphysics.com 48035239 General Physics B PHY 2049C HW Ch, 33 Problem 33.59 Problem 33.59 Part A A child's toy sits on the bottom of a swimming pool in which At what horizontal distance from the sidewall is the toy located? the water depth is d 2.1 m (Figure 1). To a child standing Express your answer with the appropriate units. at the pool edge and a distance h J 3.5 m above the pool bottom, the toy appears to be a horizontal distance y 4.2 m away from the sidewall 3.62 m toy Submit My Answers Give U incorrect, one attempt remaining, Try Again Figure 1 of 1 yoy Signed in as Albert Kishek l Help l Close Resources previous l 6 of 13 l next Continue Provide Feedback 6:10 PM le 4/19/2015

Explanation / Answer

let x is the distance from center of the pool to toy.

let theata_i is the angle of incidence,

sin(theta_i) = x/(sqrt(x^2+d^2)

= x/sqrt(x^2 + 2.1^2)

let theta_r is the angle of refraction.

sin(theta_r) = (y/2)/sqrt((y/2)^2 + (h-d)^2 )

= (4.2/2)/sqrt( (4.2/2)^2 + (3.5-2.1)^2 )

= 0.832

theta_r = sin^-1(0.832)

= 56.3 degrees

now Apply Snelss's law

n1*sin(theta_i) = n2*sin(theta_r)

1.33*x/sqrt(x^2 + 2.1^2) = 1*0.832

1.33*x = 0.832*sqrt(x^2+2.1^2)

1.7689*x^2 = 0.692*(x^2 + 4.41)

x^2*(1.7689 - 0.692) = 4.41

x = 2.02 m

so, Horizontal distance of Toy from side wall = (y/2) + x

= (4.2/2) + 2.02

= 4.12 m <<<<<<<<<<--------------Answer

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