Asked by anonymous
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?
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?
Answers
Answered by
Elena
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
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.