Hi,

I found a solution to the following problem but I don't think it is right.

Homework Statement

A ray of light enters the atmosphere of a planet. It is perpendicular to the atmosphere. When it enters the atmosphere, the index of refraction equals the index of refraction of vacuum but it increases linearly as it comes closer to the planet. The light ray has to cover a distance h in the atmosphere in order to reach the planet. When the light ray arrives at the surface of the planet, the index of refraction equals $$n_f$$. How long does it take before the light ray hits the surface of the planet?

The Attempt at a Solution

We know that the index of refraction increases linearly so its equation is of the form
$$n(x) = ax + b$$
with x being 0 at the edge of the atmosphere and being h at the surface of the planet. We know that
$$n(0) = 1$$ and
$$n(h) = n_f$$
Solving for a and b gives
$$n(x) = \frac{n_f - 1}{x} + 1$$

Since
$$v = \frac{dx}{dt}$$
then
$$dt = \frac{dx}{v}$$
Using
$$n(x) = \frac{n_f - 1}{x} + 1$$
we get
$$x = \frac{h}{n_f-1}n(x) - \frac{h}{n_f-1}$$
thus
$$dx = \frac{h}{n-1}dn$$

We also have that
$$n = \frac{c}{v}$$ so $$v = \frac{c}{n}$$
Combining this with $$dt = \frac{dx}{v}$$ gives
$$dt = \frac{h}{n_f-1}dn\frac{n}{c} = \frac{hn}{c(n_f-1)}dn$$
which gives
$$t = \int_0^{n_f} \frac{hn}{c(n_f-1)}dn = \frac{h}{c}\frac{{n_f}^2}{2(n_f-1)}$$
However, I don't think this is correct because the time must be smaller than if the light ray moves in vacuum. Since it takes $$\frac{h}{c}$$ time to cover a distance of h in vacuum, we must have that
$$\frac{{n_f}^2}{2(n_f-1)} < 1$$
in order for the solution to be correct. However, this is never when $$n \geq 1$$ so the answer must be wrong.

Where did it go wrong?

Thank you?

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The time must be greater, not smaller...

Doc Al
Mentor

The Attempt at a Solution

We know that the index of refraction increases linearly so its equation is of the form
$$n(x) = ax + b$$
with x being 0 at the edge of the atmosphere and being h at the surface of the planet. We know that
$$n(0) = 1$$ and
$$n(h) = n_f$$
Solving for a and b gives
$$n(x) = \frac{n_f - 1}{x} + 1$$
Shouldn't that be:
$$n(x) = (\frac{n_f - 1}{h})x + 1$$

Shouldn't that be:
$$n(x) = (\frac{n_f - 1}{h})x + 1$$
Yes, you're right.

The time must be greater, not smaller...
Damn I feel sooooo stupid sometimes... Must be the studying. Thanks anyway! My problem's solved then since
$$\frac{n^2}{2(n-1)} > 1$$