0
$\begingroup$

When I look at the edge of a slanted mirror it looks like its affected by color dispersion which you can see by Image 1 but, when it reflects off the flat part of the mirror it looks normal, Image 2. Can anyone describe to me what is happening?

I thought it might be due to the refraction of the glass on the front of the mirror but, wouldn't that affect the light at Image 2 also?

$\endgroup$
  • $\begingroup$ What is the light source? $\endgroup$ – PhysicsDave Jun 19 at 19:34
  • $\begingroup$ @PhysicsDave They are a halogen lamp $\endgroup$ – Rory Shaughnessy Jun 19 at 23:56
2
$\begingroup$

Most household mirrors are made with a piece of glass where the rear surface is silvered. This gives pretty good image quality while protecting the silvering from damage.

So besides any effect from the actual reflection, the image you see is produced by light that has passed through the glass (twice) as well.

Snell's law shows how the angle of light is modified by moving from air to glass. By looking at a light nearly perpendicular to the mirror, all the light is incident at an angle nearly of zero degrees and therefore the refracted light is also near zero degrees. There is almost zero difference between the wavelengths.

By looking near the edge of the tilted mirror, the incident rays have a high angle. When the rays pass into the glass, there is greater dispersion by wavelength. Blue is refracted more red is refracted less. When the reflected light leaves the glass, the refraction is undone, but the separation of the colors that was created during passage remains.

The thicker the glass and the greater the angle, the greater the effect. You wouldn't see this at all if the outer surface were mirrored.

$\endgroup$
  • $\begingroup$ Good answer, but it would be even better with a diagram/picture. $\endgroup$ – FGSUZ Jun 20 at 0:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.