Angle of rarefraction in a pool

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Homework Help Overview

The problem involves an observer viewing a coin at the bottom of a swimming pool, with specific measurements related to the observer's height and the depth of the pool. The context is centered around the concept of refraction and the calculation of angles related to light passing from water to air.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the interpretation of the distance from the observer to the coin versus the distance to the normal. There is uncertainty about the correctness of the provided answer and the application of trigonometric functions in this context.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem setup. Some guidance has been offered regarding the relationship between the observer's position and the normal line, but there is no explicit consensus on the correctness of the answer provided in the answer sheet.

Contextual Notes

Participants are working within the constraints of a practice exam scenario, which may influence their interpretations and assumptions about the problem. The presence of a diagram is noted, but its details are not discussed in depth.

famematt
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Hello,
I am doing a practice exam for i have exams coming up soon. The problem is as follows :)

Homework Statement


An observer standing on the edge of the pool views a coin on the bottom of a swimming pool.

The observers eyes are 1.50 m above the pool and she sees the coin by looking at a point from 2m from where she is standing.

The refractive index of the pool water relative to air is 1.35 and the pool is 1.50 m deep.

(1) Find the angle of rarefaction.
(a diagram is attached)

Homework Equations



Well, basically, the answer sheet says that the angle is 90 - tan-1(1.5/2)

But the 2m is the distance is from the person to the coin, not the person to the normal, is it not?

Can someone confirm to me that I am either understanding the question incorrectly or the answer in the book is wrong?

Thanks,
Matt
 

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Hello,
The two meters would indeed be to the normal. Remember, the observer is actually looking at the surface of the water where the light refracts, and hence is at the normal location.
 
Yes, but is the answer to the question wrong?
Im unsure...

It was 90-(tan-1(1.5/2))...

I THINK its wrong... I am just unsure :)
 
My initial statement was pertaining to your diagram. However, I am getting
arctan(1.5/2) as an answer.
 

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