Laser Energy Release Calculation
This solution details the calculation for the energy released during a single laser pulse in a two-level atomic system.
Given Parameters
- Energy separation between levels: $\Delta E = 2$ eV
- Number of atoms in ground state: $N_g = 4 \times 10^{20}$
- Number of atoms in excited state: $N_e = 7 \times 10^{20}$
Energy Release Calculation
The total energy released ($E$) is determined by the number of atoms transitioning and the energy difference ($\Delta E$) per atom. The calculation uses the following formula:
$ E = \frac{N_e - N_g}{2} \times \Delta E $
This formula considers half the population inversion multiplied by the energy separation.
Step-by-Step Calculation
- Determine the population inversion: $N_{inv} = N_e - N_g = 7 \times 10^{20} - 4 \times 10^{20} = 3 \times 10^{20}$ atoms.
- Calculate the effective number of atoms contributing to energy release: $N_{eff} = \frac{N_{inv}}{2} = \frac{3 \times 10^{20}}{2} = 1.5 \times 10^{20}$ atoms.
- Calculate the total energy released in electron volts (eV): $E_{eV} = N_{eff} \times \Delta E = (1.5 \times 10^{20} \text{ atoms}) \times (2 \text{ eV/atom}) = 3 \times 10^{20}$ eV.
- Convert the total energy from eV to Joules (J). Use the conversion factor $1 \text{ eV} = 1.602 \times 10^{-19}$ J: $E_J = (3 \times 10^{20}) \times (1.602 \times 10^{-19})$ J $E_J = 4.806 \times 10^1$ J $E_J = 48.06$ J.
Final Result
The calculated energy release is approximately 48.06 J. This value rounds to 48 J.
Important Questions from Optics and Diffraction
- A highly collimated laser beam with a diameter of 1 cm and wavelength 500 nm is directed from the earth's surface towards the moon (~384,000 km away from the earth). Assuming ideal diffraction limited propagation in vacuum, which of the following best estimates the diameter of the beam upon returning to the earth after reflection from an ideal reflector installed on the moon.
Three identical pinholes separated by distance $a$ along the x-axis are illuminated by a collimated monochromatic coherent beam of light (wavelength $\lambda$) as shown in the figure below.
The intensity (in arbitrary units) pattern of fringes obtained on a screen kept at distance $D$ ($D>>a$) along the z- axis is best represented by
Two coherent plane electromagnetic waves of wavelength $0.5 \ \mu\text{m}$ (both have the same amplitude and are linearly polarized along the $z$-direction) fall on the $y = 0$ plane. Their wave vectors $\mathbf{k}_1$ and $\mathbf{k}_2$ are as shown in the figure.
If the angle $\theta$ is $30^\circ$, the fringe spacing of the interference pattern produced on the plane is
The figure below describes the arrangement of slits and screens in a Young's double slit experiment. The width of the slit in $\text{S}_1$ is $a$ and the slits in $\text{S}_2$ are of negligible width.
If the wavelength of the light is $\lambda$, the value of $d$ for which the screen would be dark isA screen has two slits, each of width $w$, with their centres at a distance $2w$ apart. It is illuminated by a monochromatic plane wave travelling along the $x$-axis.
The intensity of the interference pattern, measured on a distant screen, at an angle $\theta = n\lambda/w$ to the $x$-axis is
