DIFFICULT Internal Refraction Problem?? Please Help!!?

Question

A beam of light is emitted 7.33 cm beneath the surface of a liquid and strikes the surface 6.44 cm from the point directly above the source. If total internal reflection occurs, calculate the minimum index of refraction of the liquid?

How would you solve this problem?? I only know Snell's equation and this doesn't seem to fit into it.

Answer 1

The critical angle formula is

θ = ArcSin( n1 / n2 ) where n1 < n2 are indices of refraction. Let us assume n1 = 1 for air, and we know that from the problem stated,

Tan(θ) = 6.44 / 7.33

But Tan(θ) = Sin(θ) / √ (1 - (Sin(θ))² ), which is a trig identity, so that we plug in the expression for θ, and get:

6.44 / 7.33 = (1 / n2) / √ (1 - (1 / n2)² )

and solve for n2. It comes to:

n2 = 1.515

Answer 2

Snell's law does work. This angle is called the Brewster angle

n1sin(A1)=n2sinA2

First, you've got the geometry of the beam but it gives you the angle relative to the surface

sinB=7.33/sqrt((6.44)^2+(7.33)^2) ==> B=48.7 degrees

A1=90-48.7=41.3

Internal reflection starts to happen when A2=90 degrees.

If we assume the material above the liquid is air and n2=1

we have the following use of Snell's law

x*sin(41.3)=1*sin(90) ==>x=1/sin(41.3)=1.52

Answer 3

entire inner mirrored image takes position even as mild travels from :- a million)denser to rarer medium and a couple of) perspective of occurrence is larger then serious perspective. so on your question all 3 circumstances fulfill that criteria and consequently answer is (6)all 3