By Alexander A. Kokhanovsky

Clouds have an effect on the weather of the Earth, and they're an enormous think about the elements. for that reason, their radiative houses has to be understood in nice aspect. This publication summarizes present wisdom on cloud optical houses, for instance their skill to take in, transmit, and replicate mild, which will depend on the clouds’ geometrical and microphysical features corresponding to sizes of droplets and crystals, their shapes, and buildings. moreover, difficulties on the topic of the picture move via clouds and cloud distant sensing are addressed during this e-book in nice detail.

This e-book will be an immense resource of knowledge on theoretical cloud optics for cloud physicists, meteorologists and optical engineers.

All easy rules of optics as concerning scattering of sunshine in clouds (e.g. Mie concept and radiative move) are thought of in a self constant approach. for this reason, the ebook is also an invaluable textbook to novices to the field.

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This issue is considered in detail by Rozenberg (1973). It is known from the electromagnetic theory that the intensity of energy flow at any point in the field is given by the absolute value of the Poynting’s vector P = E×H . 91) This can be shown as follows. Let us multiply Eq. 1) by H and Eq. 2) by E and subtract resulting equations from one another. 94) we have: ∇ E×H + CLOUD OPTICS 46 where we introduced the derivative 1 ∂B ∂D ∂w = H +E ∂t c ∂t ∂t . 95) This derivative can be interpreted as the rate of decrease of electric and magnetic energy stored in the volume V (Stratton, 1941).

The surface integral of Eq. 169) s Let us assume that both a particle and a surrounding medium do not absorb incident radiation. This means that the total flux passing through the surface in both directions (to and from a particle) vanishes. This is a direct consequence of the conservation of energy principle. The same applies to Ai . Then it follows: Ae = As . We see that fluxes Ae and As coincide then. Now we assume that a particle can absorb incident light. Then the value of A is not equal to zero.

The next step is to determine u i from Eq. 49). 50) and also account for identities: 1 ∂ −ikr cos θ ∂ ), (e Pn (cos θ) = −Pn1 (cos θ) , P01 (cos θ) = 0. 51) Then it follows: e−ikr cos θ sin θ cos φ = 1 (kr )2 ∞ (−i)n−1 (2n + 1) ψn (kr ) Pn1 (cos θ) cos φ. 52) CLOUD OPTICS 40 We apply the following trial solution of Eq. 53) n=1 which follows from Eq. 33) taking into account the form given by Eq. 52). Substitution of Eqs. 53) into Eq. 49) gives the following result: αn k 2 ψn (kr ) + ∂ 2 ψn (kr ) ψn (kr ) .

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