Law of Refraction with changing index of refraction

[Mnemonic]

Homework Statement

A light ray enters the atmosphere of the Earth and descends vertically to the surface a distance h = 101.2-km below. The index of refraction where the light enters the atmosphere is n = 1.00 and it increases linearly with distance to a value of n= 1.000293 at the Earth's surface.Over what time interval does the light traverse this path?

Homework Equations

v1/v2=n2/n1

v1=3e8
n1=1
v2=3e8/n2
D=distance from atmosphere barrier

The Attempt at a Solution

So the increase in n2 per m = 1/345392491

Therefore n2=(D/345392491) +1

v2=3e8/(1+(D/345392491))

So v2 changes with distance. I'm not sure where to go from here to get time.

 

[gneill]

See if you can write an expression for v as a function of altitude. That is, if y is the vertical height then v(y) = ?. It will rely on also having an expression for n(y). Then knowing that v is dy/dt you should be able to write a differential equation.

 

[Mnemonic]

See if you can write an expression for v as a function of altitude. That is, if y is the vertical height then v(y) = ?. It will rely on also having an expression for n(y). Then knowing that v is dy/dt you should be able to write a differential equation.

Thanks I got the answer!