Refraction

[Function]

Hello everyone

 

Consider a convex surface, a light ray being emitted from point O, being refracted by the surface in point P.

The light ray forms an angle alpha with the line OC, C being the centrum of the convex object. PC forms an angle beta with OC and is the normal in P, the point in which the light ray is being refracted, Q is the intersection of OC and the convex surface, OP forms an angle theta 1 with the normal, the refracted light ray forms an angle theta 2 with the normal, intersecting OC in I, the point in which the image will be projected. Gamma is the angle formed by PI and OC.

 

My syllabus states:

 

If alpha is small enough, beta, gamma, theta 1 and theta 2 will also be small and thus:

sin(theta 1) ~ theta 1 and sin(theta 2) ~ theta 2

 

After some operations, it states:

 

Expressed in radials, alpha, beta and gamma are approximately:

alpha ~ QP/OQ

beta ~ QP/OC

gamma ~ QP/QI

 

Can someone explain to me why this is the case?

 

Thanks

 

Function

[elfmotat]

 

Apologies for the sloppiness, all I had to write on was my desk!

 

So, if you can read my drawing well enough, you should be able to see that we have:

 

[math]tan \alpha = \frac{PQ}{OQ}[/math]

 

[math]tan \beta = \frac{PQ}{QC}[/math]

 

[math]tan \gamma = \frac{PQ}{QI}[/math]

 

If we expand out tanx we have:

 

[math]tan x=\frac{sinx}{cosx} = \frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-...}{1-\frac{x^2}{2}+\frac{x^4}{4!}-...}[/math]

 

Now if we let x be very small, then all powers of x will be even smaller. So if we discard all higher order terms, we can approximate tanx with:

 

[math]tan x \approx \frac{x}{1}=x[/math]

 

So when the angles in the picture are small, we have:

 

[math]\alpha \approx \frac{PQ}{OQ}[/math]

 

[math]\beta \approx \frac{PQ}{QC}[/math]

 

[math]\gamma \approx \frac{PQ}{QI}[/math]

Edited October 18, 2014 by elfmotat

[studiot]

elfmotat is pampering you - I was going to ask for a diagram.

 

 

 

So, if you can read my drawing well enough, you should be able to see that we have:

 

 

 

 

This statement probably needs further explanation.

 

We are approximating arc PQ to the perpendicular from P tp OC. The smaller alpha is the more nearly exact this is.

 

Which brings me to the sort of conditions we are talking about.

 

We assume all rays meet the curved surface near the axis OC.

That is alpha small.

Such rays are called paraxial rays.

 

These conditions are typical of eyes.

 

We are also approximating the curved surface by a circle.

This is true in spherical lenses and paraxial rays, but not true in tall thin lenses.

CP is a normal because it represents the radius of curvature at P and is therefore perpendicular to the tangent at P.

[Function]

 

Apologies for the sloppiness, all I had to write on was my desk!

 

So, if you can read my drawing well enough, you should be able to see that we have:

 

[math]tan \alpha = \frac{PQ}{OQ}[/math]

 

[math]tan \beta = \frac{PQ}{QC}[/math]

 

[math]tan \gamma = \frac{PQ}{QI}[/math]

 

If we expand out tanx we have:

 

[math]tan x=\frac{sinx}{cosx} = \frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-...}{1-\frac{x^2}{2}+\frac{x^4}{4!}-...}[/math]

 

Now if we let x be very small, then all powers of x will be even smaller. So if we discard all higher order terms, we can approximate tanx with:

 

[math]tan x \approx \frac{x}{1}=x[/math]

 

So when the angles in the picture are small, we have:

 

[math]\alpha \approx \frac{PQ}{OQ}[/math]

 

[math]\beta \approx \frac{PQ}{QC}[/math]

 

[math]\gamma \approx \frac{PQ}{QI}[/math]

 

Oh dear Lord... Thank you for your help and, obviously, your interest in this question! And, please, don't apologize... I didn't expect someone to draw this - let alone on his desk I must confess, though, that I'm glad my 'explanation' of the situation was clear enough for you to succeed in drawing almost the exact same picture as is drawn in my syllabus (forgive for not having it scanned - couldn't scan atm)

 

elfmotat is pampering you - I was going to ask for a diagram.

 

 

 

This statement probably needs further explanation.

 

We are approximating arc PQ to the perpendicular from P tp OC. The smaller alpha is the more nearly exact this is.

 

Which brings me to the sort of conditions we are talking about.

 

We assume all rays meet the curved surface near the axis OC.

That is alpha small.

Such rays are called paraxial rays.

 

These conditions are typical of eyes.

 

We are also approximating the curved surface by a circle.

This is true in spherical lenses and paraxial rays, but not true in tall thin lenses.

CP is a normal because it represents the radius of curvature at P and is therefore perpendicular to the tangent at P.

 

Thank you, too Figured too that we had to take PQ perpendicular to OC.

And yes, the eyes were indeed the subject.

Edited October 18, 2014 by Function