# Lens basic question...

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1. Jan 11, 2016

### Cozma Alex

Why does a ray of light change's it's direction as it passes the lens? It's because of the fermat principle? Is because of the index of refraction? (Why there is no index of refraction in 1/p + 1/q = 1/f ?)And if it's because of the index of refraction can we apply snell law? Once the ray passes from air to the lens it should change it's direction and it does, but when it passes from the lens to the air it should change again direction according to snell's law but it doesn't. Why?

Thanks and sorry for bad english

2. Jan 11, 2016

### Merlin3189

Both of those, I think, just different ways of looking at it.
The index of refraction is taken into account in working out the focal length f. If two lenses have the same shape, but different refractive index, they will have different focal length.
We can apply Snell's law and we do in calculating the focal length.
The ray can change direction at both surfaces, unless it crosses the surface at 90o to the surface.

3. Jan 11, 2016

### Drakkith

Staff Emeritus
The underlying reason is that light is actually an EM wave and EM waves slow down when they enter a medium other than a vacuum. When an EM wave hits a boundary between two mediums with different refractive indices at an angle, some part of the wavefront is suddenly traveling faster or slower relative to the rest of the wavefront. The change in direction is the result of this difference in speed between different areas of the wavefront.

4. Jan 13, 2016

### lychette

light changes speed as it passes from one medium into another. The change in direction occurs if the incident light is not perpendicular to the boundary between the mediums (media).
With parallel sided blocks and normal incidence there is no change in direction

5. Jan 14, 2016

### Staff: Mentor

6. Jan 14, 2016

### Merlin3189

If we already accept the notion of refraction, index of refraction and Snell's law, I can't see how it helps us to go back to EM theory, change of speed in different media, and Maxwell's equations?

It seems to me that the OP is querying why refraction and Snell's law appear not to apply consistently to lenses. I can only guess that they have seen a ray diagram where on the exit side the rays are perpendicular to the surface, or where it is drawn roughly and appears so.

In asking about the absence of refractive index in $\frac{1}{p} + \frac {1}{q} = \frac {1}{f}$
OP is simply missing the fuller equation containing the refractive index n
$\frac {1}{p} + \frac {1}{q} = (n-1) \big[ \frac {1}{R_1} -\frac{1}{R_2} + \frac {(n-1)d}{nR_1R_2} \big]≈ (n-1) \big[ \frac {1}{R_1} -\frac{1}{R_2}\big]$

So, while those more fundamental concepts are interesting, do they address the OP query?

7. Jan 14, 2016

### Andy Resnick

This is the 'thin lens equation'. In this approximation, the lens has no thickness. Conceptually, the physical lens (an object containing possibly multiple pieces of glass) has been replaced by a single planar surface assigned an optical power equivalent to the physical lens- a version of lumped analysis. The thick lens equation does involve the refractive index and curvatures of both surfaces.

8. Jan 14, 2016

### Merlin3189

If the thick lens formula "does involve the refractive index and curvatures of both surfaces", then so does the thin approximation of this formula. (See equations in my previous post.)

Just because n does not appear in this form of the formula does not mean it is not there. You might just as well say that there is no radius of curvature in the spherical mirror formula, 1/p + 1/q = 1/f

9. Jan 14, 2016

### Drakkith

Staff Emeritus
Perhaps. It's certainly true that Snell's Law applies to rays moving into and out of a lens, so I'm not sure why the OP thinks otherwise.

10. Jan 15, 2016

### Andy Resnick

The focal length of a lens/mirror is related to the optical power ; the optical power is a function of the surface curvature(s) and index of refraction relative to air (usually). Page 2 of this document has more detail: