# How Does Light Travel Through a Gradually Changing Refractive Medium?

• yoran
In summary, the conversation discusses a problem involving a ray of light entering a planet's atmosphere and the time it takes for it to reach the planet's surface. The index of refraction increases linearly and the equation for this is given. However, there is a mistake in the attempt at a solution and it is corrected by changing the equation for n(x). The time must be greater, not smaller, and the problem is solved by showing that \frac{n^2}{2(n-1)} > 1.
yoran
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?

The time must be greater, not smaller...

yoran said:

## 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:
$$n(x) = (\frac{n_f - 1}{h})x + 1$$

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

BrendanH said:
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
$$\frac{n^2}{2(n-1)} > 1$$

## 1. What is refraction?

Refraction is the bending of light as it passes through different mediums, such as air, water, or glass. This occurs because light travels at different speeds in different mediums.

## 2. How does refraction affect the appearance of objects?

Refraction can make objects appear closer or farther away than they actually are. It can also distort their shape or make them appear to be in a different location.

## 3. What causes refraction?

Refraction is caused by the change in speed of light as it passes through different mediums. This change in speed is due to the change in density of the medium.

## 4. What is the difference between refraction and reflection?

Refraction involves the bending of light as it passes through a medium, while reflection involves the bouncing of light off a surface. Refraction changes the direction of light, while reflection does not.

## 5. How is refraction used in everyday life?

Refraction is used in many everyday objects, such as eyeglasses, cameras, and microscopes. It is also used in the design of optical instruments and in the study of optics in science and engineering.

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