Question

Two coherent monochromatic light beams of intensity \( I \) and \( 4I \) are superimposed. The maximum and minimum possible intensities in the resulting beam are:

• A. 5I and I
• B. 5I and 3I
• C. 9I and I
• D. 9I and 3I

Hint:

The maximum intensity occurs when the waves are in phase, and the minimum intensity occurs when the waves are out of phase, following the principle of superposition.

Answer 1

The Correct Option is C

To solve the problem of determining the maximum and minimum possible intensities when two coherent monochromatic light beams are superimposed, we must understand the principle of superposition and interference.

When two light beams interfere, the resultant intensity depends on the phase difference between them. The resultant intensity \( I_r \) is given by:

I_r = \( I_1 + I_2 + 2\sqrt{I_1 I_2}\cos(\phi) \)

where \( I_1 \) and \( I_2 \) are the intensities of the individual beams, and \( \phi \) is the phase difference between the beams. Let's apply this for the given problem:

Given: \( I_1 = I \) and \( I_2 = 4I \)

1. **Maximum Intensity**: This occurs when the beams interfere constructively, i.e., \( \phi = 0 \). The cosine term becomes 1.

\( I_{max} = I + 4I + 2\sqrt{I \cdot 4I}\cdot 1 \)
= \( 5I + 4I \)
= \( 9I \)

2. **Minimum Intensity**: This occurs when the beams interfere destructively, i.e., \( \phi = \pi \). The cosine term becomes -1.

\( I_{min} = I + 4I + 2\sqrt{I \cdot 4I}\cdot (-1) \)
= \( 5I - 4I \)
= \( I \)

Thus, the maximum and minimum possible intensities in the resulting beam are \( 9I \) and \( I \) respectively.