Snell's Law and the Refraction of Light?

Question

A person working on the transmission of a car accidently drops a bolt into a tray of oil. The oil is 5.60 cm deep. The bolt appears to be 3.90 cm beneath the surface of the oil, when viewed from directly above. What is the index of the refraction of the oil?

Answer 1

let θi be the angle of incidence and θr be the angle of refraction. ni=1

Imagine a line going through the oil directly down to the bolt. This line forms a triangle with the refracted ray and the surface forming an angle θr. The direct ray is extended and intersects the line.
Using trig we get tan(θi)/tan(θr)=5.6/3.9

Now from Snell's law we have
ni*sin(θi)=nr*sin(θr)

and tan(θi)=sin(θi)/cos(θi)
=sin(θi)/sqrt(1-sin^2(θi))
and tan(θr)
=(ni/nr)*sin(θi)/sqrt(1-(ni/nr)^2*sin^2(θi))

dividing tan(θi)/tan(θr)
=5.6/3.9=(sin(θi)/sqrt(1-sin^2(θi))/((ni/nr)*sin(θi)/sqrt(1-(ni/nr)^2*sin^2(θi)))

which simplifies slightly to

nr/ni*sqrt(1-(ni/nr*sin(θi))^2)/sqrt(1-sin^2(θi)
ni~1
nr*sqrt(1-(1/nr*sin(θi))^2)/sqrt(1-sin^2(θi)

So, now what to do? we don't exactly know θi. But the problem says the bolt is viewed directly above so θi will be small. sine of a small angle is ~0

and nr~5.6/3.9=1.44