### Optical properties of metal

1. The optical properties of metal can be explained by considering interaction of electron and incident electromagnetic radiation.
2. Metals are opaque because incident radiation having frequencies within the visible range excite electron into unoccupied energy state above Fermi energy as shown in figure.

1. Total absorption is within a very thin outer layer usually less than 0.1 $$\mu m$$ thus so only metallic film thinner than 0.1$$\mu$$m are capable of transmitting visible lights.
2. In facts metals are opaque to all electromagnetic radiation on the low end of frequency spectrum from radio waves through infrared, visible and about middle of the ultraviolet radiation.
3. Metals are transparent to high frequency radiation i.e. X-rays and gamma rays.
4. All frequencies of visible light are absorbed by metal because of the continuously available empty electron state which permits electron transition.
5. Most of the absorbed radiation is reemitted from the surface in the form of visible light of same wavelength which appears as reflected light.
6. Aluminum and silver are two metals that exhibit this reflective behavior.
7. Cooper and gold appears red-orange and yellow respectively because some of the energy associated with light photons having short wavelength is not reemitted as visible light.
8. Why metals are transparent to high frequency X-ray and gamma ray photon?

The energy band structure of metals are such that empty and available electron state are adjacent to field states. Electron excitation from filled to empty states are possible with the absorption of electromagnetic radiation having the frequency within the visible region. The light energy is totally absorbed or reflected. None of the light energy is transmitted so the metals appear opaque.

#### Optical properties of non-metal:

Due to the electronic energy band structure non-metallic materials may be transparent to visible light. In addition to reflection and absorption refraction and transmission phenomena are also need to be considered.

#### Refraction:

The change of speed of light as it travels from on medium to another is known as refraction.

The refractive index of material or index of refraction is defined as the ratio of the velocity of light in vacuum to velocity of light in medium. It is denoted be ‘n’.

$$n=\frac{c}{v}\dotsm(1)$$Here ‘n’ is also known as degree of bending. It depends upon the wavelength of light. This effect is graphically explain by dispersion of light when it passes through a glass prism.

In medium, speed of light is given by,

$$v=\frac{1}{\sqrt{\epsilon\mu}}\dotsm(2)$$where

$$\epsilon=\epsilon_\circ \epsilon_r$$=permittivity

$$\epsilon_r$$=dielectric constant or relative permittivity

$$V=\frac{1}{\sqrt {\epsilon_r \epsilon_\circ \mu_\circ \mu_r}}$$

$$V=\frac{1}{\sqrt{\mu_\circ \epsilon_\circ}} \frac{1}{\sqrt{\mu_r \epsilon_r}}$$

$$\mu$$=permeability of material=$$\mu_\circ \mu_r$$

$$V=\frac{C}{\sqrt{\mu_r \epsilon_r}}\dotsm(3)$$

From (1) and (3)

$$\frac{c}{n}=\frac{c}{\sqrt{\mu_r \epsilon_r}}\dotsm(4)$$ for non-magnetic material(non-metals),

$$\mu_r=1$$,$$\therefore n=\sqrt{\epsilon_r}\dotsm(5)$$

Thus for the transparent material, the refractive index of material is square root of dielectric constant or relative permeability.

Equation (5) is valid for time-dependent electric field.

#### Snell’s law of refraction:

$$\frac{n}{n’}=\frac{sin \theta}{sin\theta’}\dotsm(6)$$

Where,

n=refractive index of first medium

n’=refractive index of second medium

$$\theta$$=angle of incident

$$\theta’$$=angle of refraction

#### Reflection:

The coefficient of reflection or reflectivity is defined as the ratio of intensity of reflected light to intensity of incident light. It is denoted by R.

$$R=\frac{I_R}{I_\circ}\dotsm(1)$$where

$$I_R$$=intensity of reflected light

$$I_\circ$$=intensity of incident light

The reflection or reflectivity on the case of normal incidence i.e. related to index of refraction as,

$$R=\biggl(\frac{n_2-n_1}{n_2+n_1}\biggr)^2\dotsm(2)$$where

$$n_2$$=refractive index of second medium

$$n_1$$=refractive index of first medium

If the first medium is vacuum or air then,

$$R=\biggl(\frac{n_s-1}{n_s+1}\biggr)^2\dotsm(3)$$where

$$n_s$$=refractive index of second medium

#### References:

Callister, W.D and D.G Rethwisch. Material Science and Engineering. 2nd. New Delhi: Wiley India, 2014.

Lindsay, S.M. Introduction of Nanoscience . New York : Oxford University Press, 2010.

Patton, W.J. Materials in industry . New Delhi : Prentice hall of India, 1975.

Poole, C.P. and F.J. Owens. Introduction To Nanotechnology. New Delhi: Wiley India , 2006.

Raghavan, V. Material Science and Engineering. 4th . New Delhi: Pretence-Hall of India, 2003.

Tiley, R.J.D. Understanding solids: The science of Materials. Engalnd : John wiley & Sons , 2004.

1. Some relations :

$$v=\frac{1}{\sqrt{\epsilon\mu}}$$

$$V=\frac{C}{\sqrt{\mu_r \epsilon_r}}$$

$$\mu_r=1$$,$$\therefore n=\sqrt{\epsilon_r}$$

$$\frac{n}{n’}=\frac{sin \theta}{sin\theta’}$$

$$R=\biggl(\frac{n_s-1}{n_s+1}\biggr)^2$$

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