Question 6 Thickness of semi-transparent ice: Now let’s focus on the equatorial

Question

Question 6 Thickness of semi-transparent ice: Now let’s focus on the equatorial region,where the ice layer would be expected to be the thinnest. The solar flux at the equator is about 20% higher than the global average value. Suppose that 1% of this incident sunlight made it through the ice. In that case, the ice layer must conduct out the geothermal heat flux plus the additional solar energy that passes through. How thin would the ice layer have to be in order to conduct this additional heat back out? Again use the ice thickness tool to compute the thickness in meters.

Explanation / Answer

ANSWER-

On the surface of the earth ice formation is the tedious job.As we know that ice could be form by decreasing (lapse ) in temparature.It could be either because of altitute at any where on the earth surface (average gradient in dereasing temperature with altitute in toposphere is 6.5degree centigrade per kilometer)or because of decrasing (lapse )in temperature with increasing latitudinal variation. Means at high latitude we have less than 0 degree centigrade temperature.

Ice formation because of altitudinal variation eg. at himalayan region and at pole because of latitudinal variation like arctic ice, greenland and antarctica.

At the earth surface global average solar flux value is 1050 W/m2.

At equator it is higher than 20% from 1050 W/m square.means 1050*20/100=210 W/M square.

so 1050+210=1260 W/M square at equator.

If 1% from this 1260W/M square calue means 12.60 W/M square go through the conduction and additional solar radiation so there is decrease in thickness and ice layer become thinner and thinner. As we know that maximum terristrial reflection by ice surface only so could be cause of ice layer thinning.

The SPESS program for computing the ice thickness-

In general the reflectance (R) of a thin film deposited on a substrate can be represented by a function of: the laser wavelength (0), the refractive index of the film (nf), the refractive index of the substrate (ns), the film thickness (d), the incident angle (i with respect to the normal) and the polarization of the laser light (p): R=f(0, nf , ns , d, i , p) (1) At a fixed wavelength, if we assume that there is no absorption in the film at the laser wavelength, R is a periodic function of the thickness and the period, that is the distance between two maxima or two minima in the interference curve (reflectance R versus the film thickness), is given by the equation: (2) From eq. (2) it is evident that nf , the refractive index of the film, must be known in order to measure its thickness. The amplitude of the interference curve itself depends on the refractive index of the sample, so by measuring this quantity (intensity ratio between maxima and minima) it is possible to derive the refractive index of the ice and then the thickness. In eq. (1) 0, ns , i and p are known quantities, so for a given refractive index of the film we can compute an interference curve (R versus the thickness d). We derive the refractive index of the film by using a FORTRAN code, named SPESS, that varies the nf value until the theoretical amplitude of the interference curve becomes equal to the experimental one.

Source material taken from for tool from the paper...Osservatorio Astrofisico di Catania -C.SCIRE

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