T2a3s - Physics 102 c c and in Lucite vLucite = Thus the nwater nLucite total time required to transverse the double layer is 22.8 The speed of

T2a3s - Physics 102 c c and in Lucite vLucite = Thus...

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207 Submarine 100 m air n = 1.00 water n = 1.333 θ 1 θ 1 θ 2 θ 2 210 m h 90.0 m Physics 102 Assignment 3 solutions (Spring Term) {1} 22.8 The speed of light in water is water water c v n = , and in Lucite Lucite Lucite c v n = . Thus, the total time required to transverse the double layer is 1 water Lucite water water Lucite Lucite water Lucite d d d n d n t v v c + = + = The time to travel the same distance in air is 2 water Lucite d d t c + = , so the additional time required for the double layer is ( ) ( ) 1 2 1 1 water water Lucite Lucite d n d n t t t c + ∆ = ∆ − ∆ = ( ) ( ) ( ) ( ) 2 2 8 1.00 10 m 1.333 1 0.500 10 m 1.59 1 3.00 10 m s × + × = = × 11 2.09 10 s × {2} 22.22 The angle of incidence at the water surface is 1 1 90.0 m tan 42.0 100 m θ = = ° Then, Snell’s law gives the angle of refraction as ( ) 1 1 1 2 1.333 sin42.0 sin sin sin 63.1 1.00 water air n n θ θ ° = = = ° , so the height of the building is 2 210 m 210 m tan tan63.1 h θ = = = ° 107 m
C H A P T E R 2 2 208 {3} 22.31

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