Two coherent monochromatic light beams of intensities 4I and 9I are superimposed. The difference between the maximum and minimum intensities in the resulting interference pattern is xI. The value of x is ______.
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Correct Answer: 24
Approach Solution - 1
The problem asks for the value of x, where xI represents the difference between the maximum and minimum intensities in an interference pattern formed by the superposition of two coherent monochromatic light beams with intensities 4I and 9I.
Concept Used:
When two coherent light beams with intensities \( I_1 \) and \( I_2 \) interfere, the resulting intensity at any point depends on the phase difference between the beams. The maximum and minimum possible intensities in the interference pattern are given by:
1. Maximum Intensity (\(I_{max}\)): This occurs during constructive interference, when the waves are in phase. The formula is:
\[ I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 \]2. Minimum Intensity (\(I_{min}\)): This occurs during destructive interference, when the waves are out of phase by \( \pi \) radians. The formula is:
\[ I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2 \]Step-by-Step Solution:
Step 1: Identify the intensities of the two coherent light beams.
The given intensities are:
\[ I_1 = 4I \] \[ I_2 = 9I \]Step 2: Calculate the maximum intensity (\(I_{max}\)) in the interference pattern.
First, find the square roots of the intensities:
\[ \sqrt{I_1} = \sqrt{4I} = 2\sqrt{I} \] \[ \sqrt{I_2} = \sqrt{9I} = 3\sqrt{I} \]Now, substitute these into the formula for maximum intensity:
\[ I_{max} = (2\sqrt{I} + 3\sqrt{I})^2 = (5\sqrt{I})^2 = 25I \]Step 3: Calculate the minimum intensity (\(I_{min}\)) in the interference pattern.
Using the square roots of the intensities found in the previous step, substitute them into the formula for minimum intensity:
\[ I_{min} = (3\sqrt{I} - 2\sqrt{I})^2 = (\sqrt{I})^2 = I \]Step 4: Find the difference between the maximum and minimum intensities.
The difference is \( I_{max} - I_{min} \).
\[ \text{Difference} = 25I - I = 24I \]Final Computation & Result:
The problem states that the difference between the maximum and minimum intensities is equal to xI.
Comparing our calculated difference with the given expression:
\[ 24I = xI \]By comparison, we find the value of x.
\[ x = 24 \]The value of x is 24.
Approach Solution -2
$I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$
$I_{max} = (\sqrt{4I} + \sqrt{9I})^2$
$I_{max} = (2\sqrt{I} + 3\sqrt{I})^2$
$I_{max} = (5\sqrt{I})^2 = 25I$
$I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$
$I_{min} = (\sqrt{4I} - \sqrt{9I})^2$
$I_{min} = (2\sqrt{I} - 3\sqrt{I})^2$
$I_{min} = (-\sqrt{I})^2 = I$
$I_{max} - I_{min} = 24I$
$x = 24$
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