Question

Mosquitoes are able to find a person’s vein in three stages; at a range of 50 meters the mosquito detects a CO2 plume due to breath, at 10 meters the mosquito is able to see the person in the visible range, and finally inside 20 cm the mosquito can detect the low energy light emitted from our warm veins. If the average body temperature is 37.5 ◦ C and the mosquitoes eye is sensitive enough to detect 10^16 photons per second in the infrared, estimate the surface area of the mosquitoes eye (in mm^2 ). Hint: consider the mosquito at 20 cm from a person’s arm and approximate the arm as a cylinder with a surface area of 4.5% where all radiation leaves perpendicular to the surface the total. You will need: • the average length of a person’s arm ≈ 64 cm. • the average surface area of a person ≈ 1.75 m^2 .

Answer 1

From Wein's displacement law, the wavelength of the radiation emitted will be:

for this wavelength, the energy of the photon is:

this is the energy of each photon. Number of photons received every second by the mosquito = 10¹⁶

so, Power of the radiation received = P = N x E = 10¹⁶ x 2.129 x 10^-20 = 2.129 x 10^-4 J/s = 2.129 x 10^-4 W

To find the power emitted, use Stefan-Boltzman's law,

here, A = 4.5% of surface area of the entire body = 0.045 x 1.75 m²

so,

this power is radiated radially outwards from a cylinder to a distance of R = 20 cm = 0.2 m

so, Intensity at R = 0.2 m (where the mosquito is) is:

Here, L = 64 cm = 0.64 m

so, I = 51.60 W/m²

and Intensity = power / area

so,

this is the surface area of the mosquito's eye.