Question

The concentration of photons in a uniform light beam with a wavelength of 500 nm is 1.7 x 109 mThe intensity of the beam is​

Answer 1

Answer:

We know for earlier Physics course that light consists of billions of photons and each photons carriers energy i.e E=hcλE=hcλ

So we consider n be the total number of Photon so it's Energy per unit time is =nhcλnhcλ=Power of beam = Intensity× Area =I×A=I×A

So Photon flux=Iλhc=Iλhc

= 20002.4820002.48

=806=806

Hope this helps you

Answer 2

Given:

Concentration of photons $\[ = 1.7 \times {10^9}{m^{ - 3}}\]$

Wavelength of the light beam $\[ = 500nm = 500 \times {10^{ - 9}}m\]$

To Find: Intensity of the beam

Solution:

Energy of a photon can be given as:

$\[E = \frac{{hc}}{\lambda }\]$

$\[E = \frac{{6.6 \times {{10}^{ - 34}}J \times 3 \times {{10}^8}m/s}}{{500 \times {{10}^{ - 9}}m}}\]$

$\[E = 4 \times {10^{ - 19}}J\]$

Energy density per $\[{m^3}\]$ can be given as:

$\[Energy\,density = 4 \times {10^{ - 19}}J \times 1.7 \times {10^9}{m^{ - 3}}\]$

$\[Energy\,density = 6.8 \times {10^{ - 10}}J{m^{ - 3}}\]$

The distance travelled by light beam in 1 sec is $\[{3 \times {{10}^8}m}\]$

Suppose, the cross-sectional area can be $\[A{m^2}\]$

Therefore, the volume of light passing in 1 sec can be given as:

$\[V = Area \times Length\]$

$\[V = A \times 3 \times {10^8}m\]$

Energy per second or power can be given as:

$\[Power = Energy\,density \times V\]$

$\[P = 6.8 \times {10^{ - 10}} \times A \times 3 \times {10^8}W\]$

Thus, intensity can be calculated as:

$\[Intensity = \frac{{Power}}{{Area}}\]$

$\[Intensity = 20.4 \times {10^{ - 2}}W/{m^2}\]$

$\[Intensity = 0.204\,W/{m^2}\]$

Hence, the intensity of the beam is $\[0.204\,W/{m^2}\]$.