Use the law of reflection and index of refraction to find apparent d

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Homework Help Overview

The discussion revolves around the application of the law of reflection and the index of refraction to determine the apparent depth of an object in a medium, specifically involving light behavior at the interface between water and air.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore different methods for calculating object and image distances, questioning the correctness of initial assumptions and calculations. There is a focus on applying refraction theory and considering the geometry of light paths.

Discussion Status

Some participants have provided alternative approaches and insights into the problem, suggesting the use of trigonometric relationships and the importance of accurately determining the object distance. There is an ongoing exploration of the relationships between angles and distances in the context of refraction.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the information they can share or the methods they can use. There is also a focus on ensuring accuracy in calculations before finalizing answers.

rocapp
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Homework Statement



See image

Homework Equations



n1sin(theta1) = n2sin(theta2)
1/s + 1/s' = 1/f

The Attempt at a Solution



My question is for the second part. I think I've figured out why I got it wrong, but I'm not sure.

So since the object distance (s) should be approximately 21 cm, then you apply the refraction theory to find

s'=s*n1/n2
s'=21*(1.33)
s'=~28 cm

Is this correct?
 

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No, not correct. How did you do the first part, since it's the same kind of problem? For the second part, where is the image of the fish in the mirror? That's your "object".

Once you have the correct depth of the "object", do this. Imagine a small cone of light coming from that object extending up towards the surface. If that cone has an angle θ, what will be the angle of the cone after it leaves the water? Use that and a bit of trig to figure out the apparent depth of the object. (Assume the angles involved are small.)
 
For the first part, I used

(Na/Sa) + (Nb/Sb) = (Nb - Na) / R

1.3/6 + 1/x = nb - na / R

.217 + 1/x = nb - na / infinity

1/x = .217

x = 4.61 cm
 
rocapp said:
For the first part, I used

(Na/Sa) + (Nb/Sb) = (Nb - Na) / R

1.3/6 + 1/x = nb - na / R

.217 + 1/x = nb - na / infinity

1/x = .217

x = 4.61 cm
That's an excellent way to solve the problem. But don't round off until the end and you'll get the correct answer.

The second part can be handled the same way, once you find the correct image location.
 

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