Building an X-ray Telescope

[SHISHKABOB]

Homework Statement

it gives a bit of an intro

"It is difficult to build a large surface area telescope to observe X-rays, because the index of refraction for nearly all materials at those wavelength is less than 1, which means you cannot create a reflective mirror in the normal fashion. Because the index of refraction in vacuum is larger than that in the mirror material, however, you can bounce the X-ray off the mirror using small angles via total internal reflection."

I'll start with just the first part of the problem because I don't even know where to start with the rest of it.

(a) Assuming we have built a mirror out of a substance with an index of refraction n = 1 - δ, calculate the value of δ if the mirror can only reflect X-rays hitting the mirror at 89° ≤ θ
< 90°.

Homework Equations

n₁sinθ₁ = n₂sinθ₂

The Attempt at a Solution

What I did was say the angle of total internal reflection was about 1° and set that equal to sin^-1(n₁/n₂). And if n₁ is 1 - δ and n₂ is 1 because it's vacuum (I assume) then δ = 0.982.

 

[anathema]

Hello,

Thank you for your post. Your approach is on the right track, but there are a few things to clarify and correct.

First, the angle of total internal reflection is not 1°, but rather the critical angle, which is the angle at which the refracted ray would have an angle of 90° with the normal. This is given by the equation sinθc = n2/n1, where n1 is the index of refraction of the material the light is coming from (in this case, 1 - δ) and n2 is the index of refraction of the material the light is entering (in this case, 1 for vacuum). So for X-rays hitting the mirror at 89° ≤ θ < 90°, the critical angle would be very close to 90°, which means that the index of refraction of the mirror material would need to be very close to 1.

Secondly, the equation you used, n1sinθ1 = n2sinθ2, is the Snell's law equation for refraction, not total internal reflection. Total internal reflection occurs when the incident angle is greater than the critical angle, and in this case, no refraction occurs. So the correct equation to use would be n1sinθ1 = n2sinθc, where θc is the critical angle.

Finally, to solve for δ, you would need to rearrange the equation to solve for n1. So n1 = n2sinθc/sinθ1. Plugging in the values for n2 (1) and θc (very close to 90°), and using the given value of θ1 (89°), you can solve for n1. Then, to find δ, you would subtract n1 from 1.

I hope this helps. Let me know if you have any further questions.