# Angle of rarefraction in a pool

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 im either understanding the question incorrectly or the answer in the book is wrong?

Thanks,
Matt

#### Attachments

• Untitled1.png
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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.