Optics Question -- small glass spheres to couple to fiber optic cables

[izelkay]

Homework Statement

Homework Equations

I know that yin = 0.7mm, y2 = 0mm, the index of refraction for the sphere is 1.8 and the indices of refraction for the air surrounding it is 1.

The Attempt at a Solution

Not sure how to even begin with the given information. I was thinking I could ray-transfer matrix analysis, but how would I set that up for this question?

 

[blue_leaf77]

What equation comes to mind when one is dealing with refraction between two media?

 

[izelkay]

What equation comes to mind when one is dealing with refraction between two media?

Snell's Law: n₁sinθ₁ = n₂sin₂
But how would I find θ₁?

 

[blue_leaf77]

For the first refraction, try to draw a line from the sphere center to the entrance point of the incoming ray. You may not need to calculate $\theta_1$, from the geometry it may be possible to get $\sin \theta_1$ directly in terms of the known data.

 

[izelkay]

For the first refraction, try to draw a line from the sphere center to the entrance point of the incoming ray. You may not need to calculate $\theta_1$, from the geometry it may be possible to get $\sin \theta_1$ directly in terms of the known data.

Okay I've determined that sinθ₁ = 0.7. Using Snell's law, the refracted ray inside the sphere's
θ₂ = sin^-1(0.7/1.8) = 22.8854°.

But then how do I get the angle of the incident ray when it exits the sphere? I'm just not seeing the geometry.

 

[blue_leaf77]

But then how do I get the angle of the incident ray when it exits the sphere? I'm just not seeing the geometry.

Now connect three points, the entrance point, exit point, and sphere center, what kind of triangle do you get?

 

[izelkay]

Now connect three points, the entrance point, exit point, and sphere center, what kind of triangle do you get?

An isosceles triangle. So the angle of the incident ray angle when it exits the sphere is the same as the refracted angle when it enters the sphere, θ₂ = θ₃ = 22.8854°. Trying to find the height it leaves the sphere now, I cut the triangle in two and using the refracted angle and radius, determine that the distance the ray traveled in the sphere is 1.84257 mm. The bottom angle of the isosceles angle is 180-22.8854°-22.8854° = 134.229°

If I draw a line straight from where the ray left the sphere I can determine the height. Using the geometry of the line, the left angle of this triangle should be 180° - θ₁ - 134.229° = 1.34376° Then the height is 1mm*sin(1.34376°) = 0.023451 mm.

Is that correct so far? The height seems too small to me.

 

[blue_leaf77]

I think you are correct.

 

[izelkay]

I think you are correct.

Okay, and then I can use snell's law again to determine, θ₄, the refracted exit ray's angle. Then I can say 0.023451mm*tan(θ₄) = f [mm]

 

[izelkay]

Okay, and then I can use snell's law again to determine, θ₄, the refracted exit ray's angle. Then I can say 0.023451mm*tan(θ₄) = f [mm]

Oh that's not quite right. I need to take 180 - θ₄ - upper angle of height triangle from before. Then I can say 0.023451mm*tan(θ) = f [mm], and I'm getting 0.0025 mm as the focal length.

 

[blue_leaf77]

0.0025 mm as the focal length

I guess you put one zero more than it is needed behind the decimal.

 

[izelkay]

I guess you put one zero more than it is needed behind the decimal.

Yep my mistake, 0.025 mm. Thanks for your help; much appreciated.