1
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
2
1/f=1/di+1/do
di=f•do/(do-f) =
=f•1.2f/(1.2f-f)=
=1.2f²/0.2f = 6f
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?
1 answer
3 answers