Question

The ratio of intensities at two points \( P \) and \( Q \) on the screen in a Young's double-slit experiment, where the phase difference between two waves of the same amplitude are \( \frac{\pi}{3} \) and \( \frac{\pi}{2} \), respectively, is:

• A. \( 1:3 \)
• B. \( 3:1 \)
• C. \( 3:2 \)
• D. \( 2:3 \)

Hint:

In Young's Double-Slit Experiment: - Intensity at a point is given by \( I = I_0 (1 + \cos \Delta \phi) \). - For phase differences: - \( \Delta \phi = \frac{\pi}{3} \Rightarrow I = \frac{3I_0}{2} \). - \( \Delta \phi = \frac{\pi}{2} \Rightarrow I = I_0 \). - Intensity ratio is determined using \( I_P : I_Q \).

Answer 1

The Correct Option is C

Step 1: Understanding intensity in Young's Double-Slit Experiment. The intensity at any point in an interference pattern is given by: \[ I = I_0 \left(1 + \cos \Delta \phi \right) \] where: - \( I_0 \) is the intensity of an individual wave, - \( \Delta \phi \) is the phase difference between the two waves. Step 2: Calculating intensity at \( P \) (where \( \Delta \phi = \frac{\pi}{3} \)). \[ I_P = I_0 \left(1 + \cos \frac{\pi}{3} \right) \] Since \( \cos \frac{\pi}{3} = \frac{1}{2} \), \[ I_P = I_0 \left(1 + \frac{1}{2} \right) = I_0 \times \frac{3}{2} = \frac{3I_0}{2} \] 
Step 3: Calculating intensity at \( Q \) (where \( \Delta \phi = \frac{\pi}{2} \)). \[ I_Q = I_0 \left(1 + \cos \frac{\pi}{2} \right) \] Since \( \cos \frac{\pi}{2} = 0 \), \[ I_Q = I_0 \left(1 + 0 \right) = I_0 \] 
Step 4: Finding the ratio \( I_P : I_Q \). \[ \frac{I_P}{I_Q} = \frac{\frac{3I_0}{2}}{I_0} = \frac{3}{2} \] Thus, the ratio of intensities: \[ I_P : I_Q = 3:2 \] 
Final Answer: \[ \boxed{3:2} \]