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Refraction problem

  • Thread starter roxxyroxx
  • Start date
  • #1
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Homework Statement



A penny lies on the bottom of a pool 0.75m from the edge of the pool and 1.5m below the surface. A flashlight beam is shone over the edge of the pool to illuminate the penny. At what angle to the pool deck should the flashlight be aimed?

Homework Equations



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The Attempt at a Solution


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Answers and Replies

  • #2
LowlyPion
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Homework Statement



A penny lies on the bottom of a pool 0.75m from the edge of the pool and 1.5m below the surface. A flashlight beam is shone over the edge of the pool to illuminate the penny. At what angle to the pool deck should the flashlight be aimed?
You're going to need the index of refraction aren't you?

Water is 1.33. Is that what you are supposed to use?

If you don't know it already, then maybe better look at Snell's Law.
 
  • #3
tiny-tim
Science Advisor
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Hi roxxyroxx! :wink:
A penny lies on the bottom of a pool 0.75m from the edge of the pool and 1.5m below the surface. A flashlight beam is shone over the edge of the pool to illuminate the penny. At what angle to the pool deck should the flashlight be aimed?
(I assume it means that the light enters the pool exactly at the edge?)

Use the sin/sin rule …

show us how far you get, and where you're stuck, and then we'll know how to help! :smile:
 
  • #4
48
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ok well i did sin(0.75/1.5) --> sin^-1(0.75/1.5) giving me 30 degrees so:
(1.00)(sin 30) = (1.33)(sin theta)
sin theta = 0.3759...
theta = 22 degrees
but the answer should be 54 degrees
 
  • #5
LowlyPion
Homework Helper
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ok well i did sin(0.75/1.5) --> sin^-1(0.75/1.5) giving me 30 degrees so:
(1.00)(sin 30) = (1.33)(sin theta)
sin theta = 0.3759...
theta = 22 degrees
but the answer should be 54 degrees
Isn't the angle you want from entering the water given by

tan-1(.75/1.5) = θ
 
  • #6
48
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ok so tan^-1(0.75/1.5) = 26.565...
(1.33)(sin 26.565...) = (1.00)(sin theta)
sin theta = 0.595
theta = 37 degrees
but the answer is 54 ..
 
  • #7
rl.bhat
Homework Helper
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ok so tan^-1(0.75/1.5) = 26.565...
(1.33)(sin 26.565...) = (1.00)(sin theta)
sin theta = 0.595
theta = 37 degrees
but the answer is 54 ..
This angle is the angle which the light makes with the normal to the pool. But the answer required is the angle to the pool deck.
So the required angle is 90 - theta
 
  • #8
48
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ookk thank you! >.<
 

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