Finding pool depth using Snell's law with 30° incident angle

[tomceka]

Homework Statement

A person looks at the surface of water with an angle of 30 degrees and sees a coin 0.5 m further than it really is. What is the depth (h) of the pool.
The answer I should get is around 16 cm, but I get 0.289 m. What is wrong with my solution?

Relevant Equations

Snell's law; basic trigonometry

α=30°; l=0.5 m; n1=1; n2=1.33

α+β=90°, so β=90°-30°=60°.
Using Snell's law:
sinβ/sinγ = n2/n1
sinγ≈0.651
γ≈41°.

β=γ+θ (vertical angles)
θ=60°-41°=19°

tan(θ+β)=l/h
h=l/tan(θ+γ)
h=0.5/(tan(19+41))≈0.289 m

 

[Lnewqban]

Welcome!
Why θ+β?

 

[tomceka]

Welcome!
Why θ+β?

My bad, it should be θ+γ. Although the answer is still wrong. Is there something I missed?

 

[Lnewqban]

The equation tan(θ+γ)=l/h is not correct.
The coin is resting at a point far from the vertical line.

 

[tomceka]

Do you mind explaining what do you mean by that? Is there something wrong with my diagram?

 

[kuruman]

Do you mind explaining what do you mean by that? Is there something wrong with my diagram?

The diagram is fine. It's your trig based on it that is not. Let x = distance from the coin to the vertical line. There is a right triangle that has angle γ and $x$ is opposite to it. Then the correct expression is, $\tan(\theta+\gamma)=\dfrac{l+x}{h}$.

 

[tomceka]

The diagram is fine. It's your trig based on it that is not. Let x = distance from the coin to the vertical line. There is a right triangle that has angle γ and $x$ is opposite to it. Then the correct expression is, $\tan(\theta+\gamma)=\dfrac{l+x}{h}$.

Thank you.