We are searching data for your request:

**Forums and discussions:**

**Manuals and reference books:**

**Data from registers:**

**Wait the end of the search in all databases.**

Upon completion, a link will appear to access the found materials.

Upon completion, a link will appear to access the found materials.

## Refraction at the prism - dispersion

In the figure below is the main section of a prism with a refracting angle$\gamma $ shown. The prism has the refractive index $n$ and be surrounded by air ($n\approx 1$). A ray of light falls at the angle of incidence ${\alpha}_{1}$ on the left prism surface and leaves the right prism surface under the angle of reflection after refraction twice ${\alpha}_{2}$. The angle of deflection $\delta $ can be determined from elementary geometric theorems:

the end $\delta ={\alpha}_{1}-{\beta}_{1}+{\alpha}_{2}-{\beta}_{2}$ and $\gamma ={\beta}_{1}+{\beta}_{2}$ follows $\delta ={\alpha}_{1}+{\alpha}_{2}-\gamma $. With the help of the law of refraction $sin{\alpha}_{1}=nsin{\beta}_{1}$ can be the deflection angle $\delta $ for any angle of incidence ${\alpha}_{1}$ determine: $\delta ={\alpha}_{1}-\gamma +arcsin\left(sin\gamma \cdot \sqrt{{n}^{2}-{sin}^{2}{\alpha}_{1}}-cos\gamma \cdot sin{\alpha}_{1}\right)$

To find the minimum deflection angle, the following theorem is important, which can generally be proven with the help of differential calculus:

- theorem
- In the case of a prism, the beam deflection is minimal if the entrance and exit angles are the same, i.e. if the beam passes through the prism symmetrically.

The following applies to symmetrical passage:${\alpha}_{1}={\alpha}_{2}={\scriptstyle \frac{1}{2}}\left({\delta}_{min}+\gamma \right)$ and${\beta}_{1}={\beta}_{2}={\scriptstyle \frac{1}{2}}\gamma $. With the help of the law of refraction we get for the minimum deflection angle:${\delta}_{min}=2arcsin\left(n\cdot sin\frac{\gamma}{2}\right)-\gamma $

The above equation can be used, for example, to determine the refractive index $n$ determine. With a prism with a very small refractive angle $\delta $ and symmetrical beam passage applies for the minimum deflection angle approximately: ${\delta}_{min}\approx \gamma \left(n-1\right)$

The refractive index of a substance generally depends on the wavelength of the light used. This fact is known as dispersion. Normal dispersion is used when $n$ increases with decreasing wavelength (joking phrase: the bluer the "break"). Since the prism is the deflection angle $\delta $ is determined by the refractive index, it offers the possibility of spatially separating light rays of different wavelengths, i.e. splitting them spectrally. This property is used in the prism spectrometer.

- Work order
With the help of the animation below, you can see the strength of the dispersion of the light for different prism materials.

In the case of the rainbow, the dispersion offers a particularly attractive spectacle.

- Work order
Watch the animation below to explain the natural phenomenon of the rainbow. Note that with the two colored light rays drawn in, it is not decisive whether one ray emerges from the raindrop above the other, but at what angle the respective ray falls into the observer's eye. How is it explained that the secondary arch appears paler than the main arch?