For interference of light, the intensity at maxima is given by \( I_{\text{max}} = (I_1 + I_2 + 2 \sqrt{I_1 I_2}) \), and the intensity at minima is \(I_{\text{min}} = (I_1 + I_2 2 \sqrt{I_1 I_2}) \).
Step 1: Formula for Intensity of Interference Maxima and Minima
The intensity of interference maxima and minima for two interfering beams is given by the following formulas:
For maxima: \[ I_{\text{max}} = (I_1 + I_2 + 2 \sqrt{I_1 I_2}) \]
For minima: \[ I_{\text{min}} = (I_1 + I_2 2 \sqrt{I_1 I_2}) \] Where \( I_1 \) and \( I_2 \) are the intensities of the two interfering beams.
Step 2: Substitute the Intensities Given in the Question
The ratio of the intensities is given as 9:4. So, \( I_1 = 9 \) and \( I_2 = 4 \).
For maxima: \[ I_{\text{max}} = (9 + 4 + 2 \sqrt{9 \times 4}) = 13 + 2 \times 6 = 13 + 12 = 25 \]
For minima: \[ I_{\text{min}} = (9 + 4 2 \sqrt{9 \times 4}) = 13 12 = 1 \]
Step 3: Calculate the Ratio of Intensities
The ratio of the intensity of maxima to minima is:
Final Answer: The ratio of the intensity of maxima and minima is 25:1 .
A two-level quantum system has energy eigenvalues
\( E_1 \) and \( E_2 \). A perturbing potential
\( H' = \lambda \Delta \sigma_x \) is introduced, where
\( \Delta \) is a constant having dimensions of energy,
\( \lambda \) is a small dimensionless parameter, and
\( \sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \).
The magnitudes of the first and the second order corrections to
\( E_1 \) due to \( H' \), respectively, are:
Consider a two-level system with energy states \( +\epsilon \) and \( -\epsilon \). The number of particles at \( +\epsilon \) level is \( N+ \) and the number of particles at \( -\epsilon \) level is \( N- \). The total energy of the system is \( E \) and the total number of particles is \( N = N+ + N- \). In the thermodynamic limit, the inverse of the absolute temperature of the system is:
(Given: \( \ln N! \approx N \ln N - N \))