This solution details the calculation for the energy released during a single laser pulse in a two-level atomic system.
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.
The calculated energy release is approximately 48.06 J. This value rounds to 48 J.
A monochromatic and linearly polarised light is used in a Young's double slit experiment. A linear polarizer, whose pass axis is at an angle $45^\circ$ to the polarization of the incident wave, is placed in front of one of the slits. If $I_{max}$ and $I_{min}$, respectively, denote the maximum and minimum intensities of the interference pattern on the screen, the visibility, defined as the ratio $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$ is
A 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
A film projector and microscope give equal magnification. But a film projector is not used to see living cells because
A pair of parallel glass plates separated by a distance $d$ is illuminated by white light as shown in the figure below. Also shown is the graph of the intensity of the reflected light $I$ as a function of the wavelength $\lambda$ recorded by a spectrometer.
Assuming that the interference takes place only between light reflected by the bottom surface of the top plate and the top surface of bottom plate, the distance $d$ is closest to