1
$\begingroup$

does the refractive index of a medium depends on wavelength of light passing through it ? If yes then why ?

$\endgroup$

marked as duplicate by John Rennie, jinawee, Kyle Kanos, Brandon Enright, Waffle's Crazy Peanut Mar 9 '14 at 3:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Be careful about "why" :-) . We can tell you that a given material has a certain spectral dispersion (variation in index) but exactly what causes that can get messy in a hurry. $\endgroup$ – Carl Witthoft Mar 8 '14 at 13:54
  • $\begingroup$ @CarlWitthoft youtube.com/watch?v=36GT2zI8lVA ;) $\endgroup$ – Andrestand Nov 28 '18 at 18:00
2
$\begingroup$

Refractive index of a medium depends upon the refractive index of the surroundings (when you consider the light ray passing from the surrounding into the medium), optical density, wavelength of the light and temperature.

Yeah it does depend on the the wavelength of light, because:

As wavelength of light decreases, the velocity of light decreases. Now, we know, absolute refractive index of a medium is equal to the (speed of light in air)/(speed of light in that medium), therefore if the velocity of light changes, the refractive index of the medium (for that wavelength of light) also changes.

Example: it's due to this difference in velocities that dispersion of sunlight (to form rainbow) takes place. While rainbow is formed, the different colors (with different wavelengths) bend at different angles, but as all colors of light pass through same medium, thus, refractive index of medium does depend on the wavelengths.

$\endgroup$
-1
$\begingroup$

Yes it depends on wave length it is inversely proportional to it. Let $u$ denote the refractive index, then:

$$u = c/v $$ where $c$ in the speed of light in vacuum and $v$ is the speed in a medium. And: $$v=fl$$ where $f$ denotes the frequency and $l$ the wavelength. So we get: $$ \frac{U_1}{U_2} = \frac{l_2}{l_1}$$

$\endgroup$
  • $\begingroup$ No, there are materials with negative dispersion as well. $\endgroup$ – Carl Witthoft Mar 8 '14 at 13:53

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