Question

When two light waves of equal intensity superimpose, the maximum intensity obtained is I. If the intensity of one of the waves is quadrupled, then the maximum intensity obtained is

• A. \( \frac{4I}{9} \)
• B. \( \frac{9I}{4} \)
• C. \( \frac{2I}{3} \)
• D. \( \frac{3I}{2} \)

Hint:

Remember that intensity is proportional to the square of the amplitude (\(I \propto A^2\)). Therefore, the maximum intensity formula \(I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2\) is equivalent to adding the amplitudes for constructive interference: \(A_{max} = A_1 + A_2\), and then squaring.

Answer 1

The Correct Option is B

Let the initial intensity of each of the two waves be \(I_0\).
The maximum intensity \(I_{max}\) during superposition occurs during constructive interference and is given by the formula \( I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 \).
In the first case, \(I_1 = I_2 = I_0\).
The maximum intensity is \( I = (\sqrt{I_0} + \sqrt{I_0})^2 = (2\sqrt{I_0})^2 = 4I_0 \).
From this, we can express the initial intensity \(I_0\) in terms of I: \( I_0 = \frac{I}{4} \).
In the second case, the intensity of one wave is quadrupled. Let the new intensities be \(I'_1\) and \(I'_2\).
Let \(I'_1 = 4I_1 = 4I_0\), and the other wave's intensity remains the same, \(I'_2 = I_2 = I_0\).
Now, we calculate the new maximum intensity, let's call it \(I'_{max}\).
\( I'_{max} = (\sqrt{I'_1} + \sqrt{I'_2})^2 = (\sqrt{4I_0} + \sqrt{I_0})^2 \).
\( I'_{max} = (2\sqrt{I_0} + \sqrt{I_0})^2 = (3\sqrt{I_0})^2 = 9I_0 \).
Finally, we express the new maximum intensity in terms of the original maximum intensity I.
Substitute \( I_0 = \frac{I}{4} \) into the expression for \(I'_{max}\).
\( I'_{max} = 9 \left(\frac{I}{4}\right) = \frac{9I}{4} \).