Alight detector (your eye) has an area of 1.95×10-6 m2 and absorbs 75% of the incident light, which is at wavelength 500 nm. the detector faces an isotropic source, 3.40 m from the source. if the detector absorbs photons at the rate of exactly 3.950 s-1, at what power does the emitter emit light?

Power=9.74*10^8ev/s

Explanation:

A light detector (your eye) has an area of 1.95×10-6 m2 and absorbs 75% of the incident light, which is at wavelength 500 nm. The detector faces an isotropic source, 3.40 m from the source. If the detector absorbs photons at the rate of exactly 3.950 s-1, at what power does the emitter emit light?

the rate at which proton are absorbed by the detector is related to the rate at which proton is emitted by light source

Rabs=0.75(Aabs  *Remit)/4r^2

Aabs=1.95×10-6 m2

Rabs=3.950 s-1

r=3.40 m

Remit=4*3.142*(3.4)^2*3.95

Remit=573.8/(.75*1.95×10-6 )

3.928*10^8 photon/s

enery of poton=

E=hc/lambda

E=1240ev/500nm

E=2.48ev

power of source

Remit*E=3.928*10^8*2.48

Power=9.74*10^8ev/s

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop