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1) How much longer would it take for light to travel through 1.0 km of water than 1.0 km of air?
What formula do you use? The answer is 1.1 x 10^-6 s

2) The acetate roll on overhead projector is placed at a distance 1.2f from the lens. (where f is the focal length). At what distance should the screen be placed to properly focus the image?
Why is the answer 6f?

  • Physics -

    c /v(water) = n - refractive index (for water n=1.33)
    v =c/n = 3•10^8/1.33=2.26•10^8 m/s
    t1=s/c=1000/3•10^8=3.3•10^-6 s.
    t2=s/v=1000/2.26•10^8=4.4 •10^-6 s.
    t2-t1 =(4.4 -3.3) •10^-6 = 1.1•10^-6 s
    di=f•do/(do-f) =
    =1.2f²/0.2f = 6f

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