Question

A beam of $\mathbf{500}\mathbf{}\mathbf{\text{nm}}$ light strikes a barrier in which there is a narrow single slit. At the very center of a screen beyond the single slit, ${\mathbf{\text{10}}}^{\mathbf{12}}$photons are detected per square millimeter per second.

(a) What is the intensity of the light at the center of the screen?

(b) A secood slit is now added very close to the first. How many photons will be detected per square millineter per sec and at the center of the screen now?

Short answer

(a) Intensity $\mathrm{I}=\mathbf{0}.\mathbf{3968}\frac{W}{{\mathbf{m}}^2}$

(b) Number of photons $\mathrm{n}=4\times {10}^{12}$ photons per square millimeter per second.

Step-by-step solution

Given

The energy divided by the Planck constant is the number of photons per unit time.

$\frac{\mathrm{E}}{\mathrm{h}}=\mathrm{n}.\mathrm{f}$

Where E is the energy , h is planck’s constant , n is the no of photons and f is the fequency

$\begin{array}{l}\text{Wavelength\hspace{0.17em}\hspace{0.17em}}\mathrm{\lambda }=500\mathrm{nm}\\ {10}^{12}\text{\hspace{0.17em}photonsper}\frac{{\text{mm}}^{\text{2}}}{\text{sec}}\end{array}$

Concept  used

The energy divided by the Planck constant is the number of photons per unit time.

$\frac{\mathbf{E}}{\mathbf{h}}\mathbf{=}\mathbf{n}\mathbf{.}\mathbf{f}$

Here, E is energy , h is plancks constant n is no of photons and f is frequency.

Energy of photon

a)

Trying to correlate the quantity of photons with their intensity is the main challenge. The two amounts are linearly proportionate to one another, as one might anticipate. We will apply this knowledge to solve for our unknowns in this situation and confirm the linear relationship as well. The intensity is defined as the total power per unit area.

The intensity$\left(\mathrm{I}\right)$ is defined as the total power $\left(\mathrm{P}\right)$ per unit area $\left(\mathrm{A}\right)$.

$\begin{array}{l}\mathrm{I}=\frac{\mathrm{P}}{\mathrm{A}}\\ =\frac{\mathrm{nE}}{\mathrm{At}}\\ =\frac{\mathrm{nhc}}{\mathrm{A\lambda t}}\\ =\frac{{10}^{12}\text{photons}\times 1240\text{\hspace{0.17em}nm}.\text{eV}\times 1.6\times {10}^{-19}\frac{\text{J}}{\text{eV}}}{{(1\times {10}^{-3}\text{m})}^2\times 500\text{nm}\times 1\text{s}}\end{array}$

Solve further as:

$\mathrm{I}=\mathbf{0}.\mathbf{3968}\frac{W}{{\mathbf{m}}^2}$

To obtain the total energy of the incident photons, multiply by in equation (1). Additionally, as demonstrated in issue, use the valuable fact that eVnm in equation (3).

Determine the number of photon

b)

For further information on this subject, see problem 10. However, the number of slits is directly inversely related to the amplitude of the wave, or electric field. Consequently, the intensity that is traditionally determined by the following formula,

$\mathrm{I}={\mathrm{I}}_0{\cos }^2\left(\frac{\mathrm{\varphi }}{2}\right)$is proportional to the square of the amplitude, therefore it will be four times greater, and as a result, there will be four times as many photons. Consequently, considering the linear relationship between intensity and photon count.

$\therefore \mathrm{n}=4\times {10}^{12}$ photons