Question

Consider a parallel beam of light of wavelength \[600nm\] and intensity \[100W{{m}^{-2}}\]. How many photons cross \[1c{{m}^{2}}\] area perpendicular to the beam in one second?
A. $3\times {{10}^{16}}$
B. $3\times {{10}^{14}}$
C. $3\times {{10}^{9}}$
D. $3\times {{10}^{4}}$

Answer 1

Hint: Light is made up of tiny packets of energy known as photons. The intensity of light waves is power transferred per unit area by the photons. The power can be resolved in terms of energy and time. The energy carried by the photons can be determined from the wavelength mentioned and velocity of light.

Formula used:
$n=\dfrac{IAt}{h}\times \dfrac{\lambda }{c}$

Complete step by step answer:
The intensity of the light wave is given by Power (which is rate of change of energy) per unit area. Mathematically, it is given by $I=\dfrac{E}{At}$.
As per the particle nature of light, light is composed of small packets of energy known as quanta. These quanta are also called photons.
Energy of these photons is mathematically given by, $E=n\left( h\upsilon \right)$
Where,
$n$ is the no. of photons
$h$ is the Planck’s constant
$\upsilon $is the frequency of light waves.
The quantity $h\upsilon $is together called quanta or photon.
Substituting this in the intensity of light, we have
$\begin{align}
  & I=\dfrac{n\left( h\upsilon \right)}{At}\Rightarrow n=\dfrac{IAt}{h\upsilon } \\
 & \Rightarrow n=\dfrac{IAt}{h}\times \dfrac{\lambda }{c} \\
 & \Rightarrow n=\dfrac{100\times {{10}^{-4}}\times 1}{6.626\times {{10}^{-34}}}\times \dfrac{600\times {{10}^{-9}}}{3\times {{10}^{8}}}=3.018\times {{10}^{16}} \\
\end{align}$

Therefore, the correct option is A.

Additional Information:
In the above question light is considered to be composed of particles called photons. This indicates the particle nature of the light, not wave nature.
The initial existence of photons was established by Albert Einstein in photo electric effect. This confirms the wave-particle duality of light.

Note:
You must understand that the formula used for intensity is not just specific for light, it applies for every quantity. The Energy of each photon is given by $h\upsilon $, so for $n$ photons, it will be $E=n\left( h\upsilon \right)$. The area is given in \[c{{m}^{2}}\], so we have resolved it into \[{{m}^{2}}\]. And, please note that the no. of photons do not have any units, it is just a number.