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.

Answer 2

Since the two beams are coherent, their electric field amplitudes add directly when they overlap, and intensity is proportional to the square of the amplitude. If the amplitude of the weaker beam is \( a \), then its intensity is \( I \propto a^2 = I \), and since the second beam has intensity \( 4I \), its amplitude must be \( 2a \) (because \( (2a)^2 = 4a^2 = 4I \)). The two extreme cases of overlap are when the two waves arrive perfectly in step, so the amplitudes simply add, and when they arrive perfectly out of step, so the amplitudes simply subtract. Let's use this to check each option.

1. 5I and I: Adding the amplitudes gives \( a + 2a = 3a \), and squaring gives \( 9a^2 = 9I \), not \( 5I \), so the maximum listed here is too small.
2. 5I and 3I: Neither \( 5I \) nor \( 3I \) matches \( (3a)^2=9I \) or \( (2a-a)^2=a^2=I \), so this pair does not fit the amplitude picture at all.
3. 9I and I: When the amplitudes add, \( (a+2a)^2 = 9a^2 = 9I \), matching the maximum; when they subtract, \( (2a-a)^2 = a^2 = I \), matching the minimum. Both values line up exactly with what the amplitude picture predicts.
4. 9I and 3I: The maximum of \( 9I \) is correct here, but the minimum should be \( (2a-a)^2 = a^2 = I \), not \( 3I \), so this option gets the minimum wrong.

Working from the amplitudes directly, rather than the resultant-intensity formula, still gives a maximum of \( 9a^2 \) and a minimum of \( a^2 \), which in terms of \( I \) are \( 9I \) and \( I \).

Therefore, the correct answer is 9I and I.