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# Optics Calculators

## Refraction

Refraction and reflection both involve changing the direction of light. In refraction, the light continues into the material, but not necessarily in its original direction. In reflection, light does not pass into the material but returns to the material it originated in. Reflection is simpler than refraction, in that the angle of incidence upon a material is the same as the angle of reflection backwards.

The change of angle in refraction occurs because light changes speed (i.e., "phase velocity") depending on the type of medium it is travelling through. From conservation of energy, the frequency, $$f=E/h$$ cannot change. However, $$f = v/\lambda$$, so that if the velocity changes, then so must the wavelength. The momentum of the light is conserved parallel to the plane, such that $$p_1 \sin(\theta_1)\ = p_2 \sin(\theta_2)$$ and because $$p = h/\lambda$$ we see that $$\lambda_2 \sin(\theta_1)\ = \lambda_1 \sin(\theta_2)$$. Thus, if the wavelength changes, because the speed changes, then so must the angle!

## Snell's Law

At the interface between two materials with different refractive indices the direction of light changes.

Snell's law is the equation used to calculate refraction:

$$\frac{\sin\theta_1}{\sin\theta_2} = \frac{v_1}{v_2} = \frac{n_2}{n_1}$$

This equation relates the angles of incidence, $$\theta_1$$, and refraction, $$\theta_2$$, to the refractive indices, $$n_1$$ and $$n_2$$, of the materials the light is passing through, and to the velocity of light, $$v_1$$ and $$v_2$$, in those materials.

If you see "invalid input" that means that the combinations of fields is not physically possible.

Because the refractive index is wavelength dependent, different colors of light get bent to slightly different angles. This is how prisms and rainbows work; the spreading of white light when passing through the glass or waterdrop.

## Light Refraction

Change the values on the left to see how it affects the angles of incidence and refraction.

If $$n_1$$ > $$n_2$$ there is a critical angle of incidence for which there is no refraction. That is, where the angle of refraction is 90°, along the material interface. For all greater angles, there is "total internal reflection". To get the critical angle, input $$\theta_2$$=90° and the desired indices of refraction ($$n_1$$ > $$n_2$$), then click "Calc" next to $$\theta_1$$ to get $$\theta_1 = \theta_c$$.

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A small amount of the incident light is reflected rather than refracted. For some angle/index combinations, you should be able to see the reflect light ray in the above demo. The Fresnel equations below can be used to calculate the fraction of light that is transmitted or reflected.

## Fresnel Equations

The following are equations governing how much light is refracted and how much is reflected when light passes an interface of different materials. They only hold for plane waves on flat, homogeneous, non-magnetic surfaces. Reflectance is the fraction of incident electromagnetic power that is reflected at the interace, while transmission is the fraction of power that is transmitted at the interface.

Reflectance for s-polarized light, where the electric field is perpendicular to the plane of the interface:

$$R_\mathrm{s} = \left|\frac{n_1\cos\theta_{\mathrm{i}}-n_2\cos\theta_{\mathrm{t}}}{n_1\cos\theta_{\mathrm{i}}+n_2\cos\theta_{\mathrm{t}}}\right|^2$$ $$=\left|\frac{n_1\cos\theta_{\mathrm{i}}-n_2\sqrt{1-\left(\frac{n_1}{n_2} \sin\theta_{\mathrm{i}}\right)^2}}{n_1\cos\theta_{\mathrm{i}}+n_2\sqrt{1-\left(\frac{n_1}{n_2} \sin\theta_{\mathrm{i}}\right)^2}}\right|^2$$

Reflectance for p-polarized light, where the electric field is parallel to the plane of the interface:

$$R_\mathrm{p} = \left|\frac{n_1\cos\theta_{\mathrm{t}}-n_2\cos\theta_{\mathrm{i}}}{n_1\cos\theta_{\mathrm{t}}+n_2\cos\theta_{\mathrm{i}}}\right|^2$$ $$=\left|\frac{n_1\sqrt{1-\left(\frac{n_1}{n_2} \sin\theta_{\mathrm{i}}\right)^2}-n_2\cos\theta_{\mathrm{i}}}{n_1\sqrt{1-\left(\frac{n_1}{n_2} \sin\theta_{\mathrm{i}}\right)^2}+n_2\cos\theta_{\mathrm{i}}}\right|^2$$

By conservation of energy, the corresponding transmission coefficients are $$T_s = 1- R_s$$ and $$T_p = 1-R_p$$

For unpolarized light, the two polarizations can be combined: $$R_u = (R_s+R_p) / 2$$

## Fresnel Equations in action

Change the values of $$n_1$$ and $$n_2$$ to see how it affects the reflection coefficients, Brewster's angle, $$\theta_B$$, and the critical angle $$\theta_C$$.

Brewster's angle, $$\theta_B$$, is the angle of incidence at which polarized will be full transmitted with zero reflectance. The angle depends on the ratio of the indices of refraction of the materials: $$\theta_B = \arctan(n_1/n_2)$$. Unpolarized light incident at this angle will be partially reflected with perfect polarization. The transmitted light is only partially polarized.

Example: Polarized sunglasses use Brewster's angle to reduce glare. For instance, light bouncing off some horizontal surfaces (water, pavement) above $$\theta_B$$ will become horizontally polarized. Sunglasses can be made of vertically polarizing material, thus not allowing through the horizontally polarized light, and reducing glare.

If $$n_1 \gt n_2$$, there is a critical angle of incidence, $$\theta_C = \arcsin(n_2/n_1)$$ for which there is no refraction and instead there is only "total internal reflection".

Example 1: Diamond has a high index of refraction (2.419) and thus a small critical angle in air. Light that enters will undergo total internal reflection several times before leaving the material. A diamond can be specially cut to use this effect to maximize sparkle.

Example 2: Fiber optic cables typically have a transparent core with n ~ 1.62 and an opaque coating with n ~ 1.52. Light that enters the fiber at a small enough angle will be reflected almost entirely back into the core as it moves along the fiber, with minimal signal loss.

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