QJEEM24Ph9099: JEE Main 2024 Physics - Class 12 JEEM24Ph9099 Solution: Two coherent monochromatic light beams of intensities I and are …

When two coherent light beams interfere, the resulting intensity at a point depends on the principle of superposition and is related to their amplitudes. Let the amplitude of the first light beam be A, then its intensity, which is proportional to the square of its amplitude, is given as $I \propto A^2$ . Since intensity is directly proportional to the square of amplitude, for the second beam with intensity $4I$ , its amplitude would be $2A$ , as $4I \propto (2A)^2$ .

The maximum intensity ( $I_{max}$ ) occurs when the two beams are in phase and their amplitudes add up constructively, which can be represented as:

$I_{max} \propto (A + 2A)^2 = (3A)^2 = 9A^2$

Given that $I \propto A^2$ , substituting this relation to express $I_{max}$ in terms of $I$ , we get:

$I_{max} = 9I$

The minimum intensity ( $I_{min}$ ) occurs when the two beams are completely out of phase, leading their amplitudes to subtract destructively, thus

$I_{min} \propto (2A - A)^2 = A^2$

Again, using the fact that $I \propto A^2$ , we find that:

$I_{min} = I$

Now, the difference between the maximum and minimum intensities in the resulting beam is:

$xI = I_{max} - I_{min} = 9I - I$

$xI = 8I$

Therefore, the value of $x$ is 8.