Ray tracing with matrices

[frankR]

Here is the problem:

A glass sphere with a diameter of 5cm has a scratch on its surface. When the scratch is viewed through the glass from a position directly opposite, where is the virtual image of the scratch, and its magnification? The glass has an index of refraction n=1.50. Explain the result.

[Answer: s'=-10cm; 3x]




I've constructed a system of matrices that looks like this:

M₃*M₂*M₁

M₁ is for the first refraction of the scratch at the sphere's front surface.
M₂ is for the translation of the ray through the glass.
M₃ is for the second refraction through the second surface of the glass where the viewer is located.


When I multiply the matrices together I get these equations.

y_f=-2/3y_o + 10/3α_o

α_f=-4/15y_o + 1/3α_o


I do not understand how I use the geometry of the ray to locate an image of a scratch. What am I missing in all of this? Did I use the matrix equation properly? What do I do, I'm extremely stumped?

Thanks for any help.

Frank

 

[frankR]

Nevermind I fiquered it out. Not sure why it's right, but I discovered how to find the correct answer using, a different system of matrices, the correct system. I didn't need to include a diffraction for the scratched surface (apparently). If someone could explain to me why not, I would be greatful.

Thanks

 

[M98Ranger]

Hi Frank,

Ray tracing with matrices can definitely be confusing at first, so let's break down the problem step by step.

First, let's define the variables we will be using:
- s: distance from the center of the sphere to the scratch
- s': distance from the center of the sphere to the virtual image of the scratch
- d: diameter of the sphere
- θ: angle of incidence (measured from the normal)
- φ: angle of refraction (measured from the normal)
- n: index of refraction

Now, let's start with the first refraction at the front surface of the sphere. We can use Snell's law to relate the angles of incidence and refraction:
n*sin(θ) = sin(φ)

Next, we can use the geometry of the ray to find the distance s' from the center of the sphere to the virtual image of the scratch. Since the ray is refracted at the front surface of the sphere, the distance s' is equal to the distance s. So we have:
s' = s

Now, let's move on to the second refraction at the back surface of the sphere where the viewer is located. Again, we can use Snell's law to relate the angles of incidence and refraction:
n*sin(φ) = sin(θ')

Using the geometry of the ray again, we can find the distance from the center of the sphere to the point where the ray intersects the back surface. Let's call this distance x. We know that the total distance from the center of the sphere to the viewer is d/2, and the distance from the center of the sphere to the point where the ray intersects the back surface is x, so the distance from the point of intersection to the viewer is d/2 - x.

Now, using similar triangles, we can relate the distances s and d/2 - x with the angles θ and θ':
s/(d/2 - x) = tan(θ')

Finally, we can use the equation for magnification to find the magnification of the virtual image of the scratch:
M = -s'/s = -s/(d/2 - x)

Plugging in our values for s' and s, we get:
M = -s/(d/2 - x)

Now, to tie all of this together with matrices, we can use