Fish has gone missing in a tank

[Les talons]

Homework Statement

At the aquarium where you work, a fish has gone missing in a 8.00mdeep, 8.80m diameter cylindrical tank. You shine a flashlight in from the top edge of the tank, hoping to see if the missing fish is on the bottom. What’s the smallest angle your flashlight beam can make with the horizontal if it’s to illuminate the bottom?

Homework Equations

n₁sinΘ₁ = n₂sinΘ₂
index of refraction of water: n_w = 1.333
index of refraction of air: n_a = 1.000293

The Attempt at a Solution

I thought that the minimum angle would be when the beam hit the edge of the tank and formed a right triangle so the angle in the water would be:
Θ₂ = arctan(8/8.8)
n₂ = n_w = 1.333
n₁ = n_a = 1.000293
Θ₁ = arcsin[(1.333/1.000293)sin(arctan(8/8.8)] = 63.69°
So then
1.333sin(42.27°) = 1.000293sin(63.69°)
But this is wrong. I also tried to subtract it from 90° but that is wrong as well, to measure from the vertical instead. Thanks in advance.

 

[Bystander]

1. (snip) 8.00mdeep, 8.80m diameter cylindrical tank.(snip)

Θ₂ = arctan(8/8.8)
(snip).

Sketch it out, and think over the trig definitions.

 

[Les talons]

Okay, I am confused what is meant by illuminate the bottom. We model the wave fronts of the light source as a ray, so to illuminate the bottom, will this ray have to reach the center of the bottom of the tank, or just an edge of the bottom of the tank where it meets the wall?

 

[Bystander]

Edge-wall boundary is good enough. Think, what is the definition of θ₂ in Snell's law?

 

[Les talons]

Hmm, Θ₂ in Snell's Law is the angle of refraction of the light ray measured with respect to the normal to the surface of the refracting medium, or "to the vertical." Ah, so I swapped the sides of the right triangle (which is what you had hinted at in outlining), and I was getting the cotangent of the angle Θ₂. Θ₂ = tangent (opp./adj.) = tan(diameter/height) = tan(8.8/8). So then I can apply Snell's Law to get:
1.333sin(arctan(8.8/8)) = 1.000293sinΘ₁
So if Snell's Law can be used to find Θ₁, the angle of incidence of the incident light ray with respect to the normal to the surface, I can take the complement of the angle to get the angle measured with respect to the horizontal, right?
I can also use this identity that I came across:
sin(arctan(x)) = x/√(1 +x^2)

 

[Bystander]

Good hunting --- we'll hope "Nemo" just ducked out for a burger.

 

[Les talons]

:p I will be finding Dory now... Thanks a bunch for the posts! I need to be more careful with what the formulas mean in words, so then I will be less likely to apply them blindly and incorrectly.