Step 1: Using the formula for intensity in a Young's double slit experiment.
The intensity at any point in Young's double slit experiment is given by: \[ I_{\text{res}} = 4I_0 \cos^2 \left( \frac{\theta}{2} \right) \] Where \( I_0 \) is the maximum intensity, and \( \theta \) is the phase difference. For the first point \( P \), where \( \theta = \frac{\pi}{3} \): \[ I_1 = 4I_0 \cos^2 \left( \frac{\pi}{6} \right) = 4I_0 \left( \frac{\sqrt{3}}{2} \right)^2 = 4I_0 \times \frac{3}{4} = 3I_0 \] For the second point \( Q \), where \( \theta = \frac{\pi}{2} \): \[ I_2 = 4I_0 \cos^2 \left( \frac{\pi}{4} \right) = 4I_0 \left( \frac{1}{\sqrt{2}} \right)^2 = 4I_0 \times \frac{1}{2} = 2I_0 \] Thus, the ratio of intensities is: \[ \frac{I_1}{I_2} = \frac{3I_0}{2I_0} = \frac{3}{2} \]
Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12. The current in Amperes used for the given electrolysis is ….. (Nearest integer).
If the system of equations \[ x + 2y - 3z = 2, \quad 2x + \lambda y + 5z = 5, \quad 14x + 3y + \mu z = 33 \] has infinitely many solutions, then \( \lambda + \mu \) is equal to:}