A source emits electromagnetic waves at 500 Joules per second with a wavelength

Question

A source emits electromagnetic waves at 500 Joules per second with a wavelength of 500nm equally in all directions in air.

(a) What is the frequency of this light?

(b) What is the approximate color of this light?

(c) If you add 1000nm to the wavelength of this light, where in the electromagnetic spectrum will it fall.

(d) The intensity of the light is measured at a distance of 200m and again at 500nm. What is the ratio of the light's intensity at these two locations?

(e) If you added 1000nm to the wavelength of this light and then measured the intensity of the light at 200m and again at 500m. What would be the ratio of the light's intensity at these two location?

Explanation / Answer

Given

Power P=100 W

speed of light c=3*108 m/s

wavelength lambda=500 nm

a)

Frequency of light

f=c/lambda =3*108/(500*10-9)

f=6*1014 Hz

b)

Color is Cyan for wavelength interval 490 nm to 520 nm ,so given wavelength is 500 nm

c)

Now Wavelength

lambda =1000+500 =1500 nm

Electromagnetic spectrum will fall in infrared region for wavelength>1000 nm

d)

Intensity is given by

I=P/4pi*r2

For r=200 m

I200 m =500/4pi*2002 =9.95*10-4 W/m2

For r=500 nm

I500 nm =500/4pi*(500*10-9)2 =1.6*1014 W/m2

SO ratio is

I200/I500nm =9.95*10-4/1.6*1014 =6.25*10-18

e)

at r=500 m

I500m =500/4pi*5002 =1.59*10-4

So ratio is

I200/I500 =9.95*10-4/1.59*10-4=6.25

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