a level physics wave questions

unclebutcher SG

ok, QN 1.

A satellite passing the planet Neptune communicates with its controller on earth using a microwave transmitter with output power 22.0W and wavelength 79600 (miu meters). Neptune is 4.35*10^12 m from earth at the time when communication took place.

assuming that the power transmitted by the satellite is radiated uniformly in all directions, calculate the power received on the earth by a dish aerial with effective area 260m^2.

actual power received at dish aerial is 1.2*10^-15 W.

1.)suggest why the actual power received is greated than calculated above.

2.) Calculate the actual rate at which photons of microwaves arrive at the dish aerial.

QN2.)

intensity of a wave depends on the amplitude. the intensity is also proportional to the square of the frequency.

a wave has frequency 3.0HZ amplitude 1.5cm and intensity I.

what is the intensity of a similar wave of freq. 6.0HZ and amplitude 0.5cm???

thanks. my physics abit noob Embarassed

airgrinder SG

I only know F=ma

unclebutcher SG

dot dot dot that one sec one students also know

eagle SG

I'm not so sure for the first part... But this is what I would do.

Power received = 260 / (4 * pi * Rneptune^2) * 22 = 2.41 * 10^-23 W

1) Why is it greater? Not sure about this...

2) E=hf=hc/lambda for photons mah...

1 second = 1.2*10^-15 J
Hence, got 1.2*10^-15/hf = 480M photons/s

unclebutcher SG

hey thanks eagle

eagle SG

This is maths...

intensity directly proportional to amplitude => 1/3 I

intensity proportional to square of freq => 1/3 I * (6/3)^2 = 4/3 I

eagle SG

teach me if u know 1)

unclebutcher SG

why is P= area/ what is 4pi Rneptune^2??? i dont get the formula...

what is hf and hc/lambda?

eagle SG

wrong leh...

F = d (mv)/dt

When mass is constant, it reduces to F=ma

teraexa SG

If Mr Eagle haven't answer by tonite, then I answer tomorrow. Damn sleepy now.

Will give a look tmr!

tanjun SG

for the 1st qn, is it because of the presence of cosmic rays that slightly affects the readings?

unclebutcher SG

weird the answer is 4/9I

eagle SG

I not sure also... I imagined a giant spherical area from the satellite till it reaches Earth. Then I took the fraction of the aerial disc's surface area over the the spherical area, then multiply by 22W

Energy of a photon is hf, which is planck's constant * frequency

since c = frequency * lambda (from v = f*lambda),
blah blah blah... u shud understand Wink

rathcycle SG

i sec 4 i dunno Rolling Eyes

eagle SG

oops... Intensity proportional to square of amplitude... You should be able to solve it now...

A bit sleepy... haha

unclebutcher SG

your answer is right.....

eagle SG

OH! you are almost there... I would say rays by other sources, including the suns, the stars, etc etc... They all may contain microwaves, which will increase the total power arriving at the disc...

Good job!

unclebutcher SG

hmm can explain further? i get the multiply by 22w part but i dun get how you arrive at the fraction of (260)/ 4 PI R(neptune)^2

eagle SG

Imagine a giant sphere, with the centre at the satellite, the surface at Earth. The rays from the satellite is directed at all parts.

The aerial disc will just be a small part of this giant sphere... So we take by proportion... the amount of energy that reached that particular area...

unclebutcher SG

ok i get it now, but how come got 4piR^2? shouldnt it be simply piR^2? how come got extra 4? sorry my physics damn noob