Dispersion/total internal reflection

[cerberus9]

Homework Statement

Consider a common mirage formed by superheated air just above a roadway. A truck driver whose eyes are 2.00 m above the road, where n=1.0003 looks forward. She has the illusion of seeing a patch of water ahead on the road, where her line of sight makes an angle of 1.20 degrees below the horizontal. Find the index of refraction of the air just above the road surface. (Hint: Treat this as a problem one involving total internal reflection)

Homework Equations

[tex]\Theta[/tex]_c=sin^-1(n₁/n₂)

n₁sin[tex]\Theta[/tex]₁=n₂sin[tex]\Theta[/tex]₂

The Attempt at a Solution

So since the textbook said to treat it as a problem with total internal reflection, i figured that I would solve for [tex]\Theta[/tex]_c and so I did

[tex]\Theta[/tex]_c=sin^-1(n₁/n₂)
[tex]\Theta[/tex]_c=sin^-1(1/1.0003)
[tex]\Theta[/tex]_c= 88.6

and I just have absolutely no idea what to do from here.

 

[rl.bhat]

n1*sin(theta 1) = n2*sin(theta 2)
In the problem n1, theta 1 and theta 2 is known. Find n2.

 

[kerimek]

I would first clarify the problem by defining the variables and providing context. In this scenario, the index of refraction of the air just above the road surface is unknown, and the angle of incidence and angle of refraction are given. The problem also mentions a mirage caused by superheated air, which is important to consider in understanding the phenomenon.

Next, I would approach the problem by using the Snell's law equation, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the indices of refraction of the two media. In this case, the two media are air and the superheated air above the roadway. The angle of incidence is the angle at which the truck driver's line of sight meets the road surface, and the angle of refraction is the angle at which the light is bent as it passes through the boundary between the two media.

Using this equation, we can rewrite it as n1sin\Theta1=n2sin\Theta2, where n1 is the index of refraction of air and n2 is the index of refraction of the superheated air. We know that the angle of incidence is 1.20 degrees and the angle of refraction is 90 degrees (since the light is refracted along the surface of the road). Substituting these values into the equation, we get:

n1sin(1.20)=n2sin(90)

Solving for n2, we get:

n2=n1sin(1.20)/sin(90)

Now, we need to find the value of n1. We can use the given information that the truck driver's eyes are 2.00 m above the road, where n=1.0003. This value of n is the index of refraction of air at standard temperature and pressure. We can use this to find the index of refraction of air at the height of the truck driver's eyes, which is:

n1=n(2.00/2.00)=1.0003

Substituting this value into the equation for n2, we get:

n2=1.0003sin(1.20)/sin(90)

Solving for n2, we get:

n2=1.0003(0.0209)/1=0.0209

Therefore, the index of