In optics, the refractive index (also known as refraction index or index of refraction) of a material is a dimensionless number that describes how fast light travels through the material. It is defined as
n
=
c
v
,
{\displaystyle n={\frac {c}{v}},}
where c is the speed of light in vacuum and v is the phase velocity of light in the medium. For example, the refractive index of water is 1.333, meaning that light travels 1.333 times slower in water than in a vacuum. Increasing the refractive index corresponds to decreasing the speed of light in the material.
The refractive index determines how much the path of light is bent, or refracted, when entering a material. This is described by Snell's law of refraction, n1 sinθ1 = n2 sinθ2,
where θ1 and θ2 are the angles of incidence and refraction, respectively, of a ray crossing the interface between two media with refractive indices n1 and n2. The refractive indices also determine the amount of light that is reflected when reaching the interface, as well as the critical angle for total internal reflection, their intensity (Fresnel's equations) and Brewster's angle.The refractive index can be seen as the factor by which the speed and the wavelength of the radiation are reduced with respect to their vacuum values: the speed of light in a medium is v = c/n, and similarly the wavelength in that medium is λ = λ0/n, where λ0 is the wavelength of that light in vacuum. This implies that vacuum has a refractive index of 1, and that the frequency (f = v/λ) of the wave is not affected by the refractive index. As a result, the perceived color of the refracted light to a human eye which depends on the frequency is not affected by the refraction or the refractive index of the medium.
The refractive index varies with wavelength, this causes white light to split into constituent colors when refracted. This is called dispersion. It can be observed in prisms and rainbows, and as chromatic aberration in lenses. Light propagation in absorbing materials can be described using a complex-valued refractive index. The imaginary part then handles the attenuation, while the real part accounts for refraction. For most materials the refractive index changes with wavelength by several percent across the visible spectrum. Nevertheless, refractive indices for materials are commonly reported using a single value for n, typically measured at 633 nm.
The concept of refractive index applies within the full electromagnetic spectrum, from X-rays to radio waves. It can also be applied to wave phenomena such as sound. In this case, the speed of sound is used instead of that of light, and a reference medium other than vacuum must be chosen.In terms of eye glasses, a lens with a high refractive index will be lighter and will have thinner edges than its conventional "low" index counterpart. Such lenses are generally more expensive to manufacture than conventional ones.
"An electromagnetic radiation has a frequency of 5 x 10^14Hz
a) (wave in vacuum = 600nm)
b) (wave in water = 440 nm)
c) what is the index of refraction of a medium in which the speed of this radiation is 2.54 x 10^8 m/s?"
so given: c = 2.54 x 10^8 m/s
f = 5 x 10^14Hz...
I can't seem to figure these out guys...maybe I'm missing a few equations too.
1. If light of wavelength 600nm in air enters a medium with index of refraction 1.50, its wavelength in this new medium is...
A. unchanged
B. 900nm
C. 400nm
D. 602nm
---I know that n=c/v...but I don't...
When constructing a Rayleigh Refractometer the formula for the refractive index of a gas at pressure P and temperature T is:
mu(P,T) - 1 = (gamma) P/T
where,
mu(P,T) = refractive index as a function of pressure and temperature
and
gamma = [n(lambda)Ta]/[L(deltaP)]
where,
n = fringe...
When constructing a Rayleigh Refractometer the formula for the refractive index of a gas at pressure P and temperature T is:
mu(P,T) - 1 = (gamma) P/T
where,
mu(P,T) = refractive index as a function of pressure and temperature
and
gamma = [n(lambda)Ta]/[L(deltaP)]
where,
n = fringe...
Hey , I need some indication to solve this problem:
A ray light impinges at a 60 degres angle of incidence on a glass pane of thickness 5mm and index of refraction of 1.54.
the ligth is reflected by a mirror that touches the back of the pane. By how much is the beam displaced compared with...
Given any index of refraction, how would I find a wavelength in that medium? Should I use frequency= c/wavelength? For example, if a piece of glass (medium) has a index of refraction 1.12, what's the wavelength?
I am having a little bit of trouble finding the speed of light in a given medium when you are only supplied with the index of refraction. For example, I was given that n=1.36 in ethanol, and I had to find the speed of sound in ethanol. I know the equation for the index of refraction is n=c/v...
imagine there is a spherical glass object with index of refraction N
is that possible that for some N and angle of incidence A, the light will be total internal reflected forever. In other word, the light will travel within the object forever?
i have been trying to proof it. but i don't know...
Hi guys.
I'm doing this science fair project and I needed to know if there is a substance (any substance!) whose index of refraction can be changed/regulated by applying an electric current (or any other action) to it.
I appreciate all help. Thanks!
given is a graph of n(lambda) vs lambda where n is the index of refraction
N(1000) =1.45
Estimate Vphase and group velocity using the above info.
i know that n = \frac{c}{v_{\phi}} = \frac{ck}{\omega}
i can't simply susbtitute into that above relation because the lambda given is that...
I have input rays into a glass plate and output rays for this glass plate also. How can I prove that they are indeed parallel. I'm thinking all I need to do is extend the inout and output rays, and indicate that they are both 180 deg. Visually, they will appear parallel anyways.
I'm a little confused. My source says the dielectric constant for water is 80. Then I have the equation that gives the index of refraction as n=\sqrt{\epsilon_r} (since it isn't very magnetic). But the index of refraction for water is 1.33. What am I missing?
If
Sin(theta r)/Sin(theta i) is reversed from Sin(theta i)/Sin(theta r) = n, what does this mean? I'm quite confused since I'm doing a physics lab, and the class was told to make a graph where we measured angles with polar paper, pins and plexyglass. Could it possibly the index of refraction...
In this figure the path of the light passes from air to glass. Calculate the index of refraction
n1 = air = 1
n2 = glass = ?
Sin theta 1 = 30degrees
sin theat 2 is = 20 degrees
formula used n2 = n1*sin theta 1 / sin theta 2
sin of 30/ sin of 20 = n2 = 1.46 is this correct
hope the...
Question: One of the beams of an interferometer, as seen in the figure below, passes through a small glass container containing a cavity D = 1.40 cm deep.
When a gas is allowed to slowly fill the container, a total of 230 dark fringes are counted to move past a reference line. The light used...