Physics Q41 – JEE-MAIN 2025

Two monochromatic light beams have intensities in the ratio

1 : 9

. An interference pattern is obtained by these beams. The ratio of the intensities of maximum to minimum is

Clue 1 of 3: Understand the relationship between intensity and amplitude

Intensity of a light wave is proportional to the square of its amplitude ( $I \propto A^{2}$ ).

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Ninja Strategy•Amplitude-Intensity Relationship

Recognize that intensity is proportional to the square of amplitude, and maximum/minimum intensities depend on the square of the sum/difference of amplitudes. This immediately rules out options that are simple amplitude ratios or initial intensity ratios.

🧠 Full Solution Path

Step 1: Relate intensity ratio to amplitude ratio✦ Active

The intensity of a light wave is proportional to the square of its amplitude. Given the ratio of intensities $I_{1} : I_{2} = 1 : 9$ , we can find the ratio of amplitudes $A_{1} : A_{2}$ .

\frac{I_{1}}{I_{2}} = \frac{A_{1}^{2}}{A_{2}^{2}} ⟹ \frac{A_{1}}{A_{2}} = \sqrt{\frac{I_{1}}{I_{2}}} = \sqrt{\frac{1}{9}} = \frac{1}{3}

Let $A_{1} = A$ and $A_{2} = 3 A$ for simplicity.

Step 2: Calculate maximum and minimum intensities○ Expand

For constructive interference, the amplitudes add, resulting in maximum intensity. For destructive interference, they subtract, resulting in minimum intensity.

I_{m a x} = (A_{1} + A_{2})^{2} = (A + 3 A)^{2} = (4 A)^{2} = 16 A^{2}

I_{m i n} = (A_{1} - A_{2})^{2} = (A - 3 A)^{2} = (- 2 A)^{2} = 4 A^{2}

Step 3: Determine the ratio of maximum to minimum intensities○ Expand

Divide the maximum intensity by the minimum intensity to find the required ratio.

\frac{I_{m a x}}{I_{m i n}} = \frac{16 A^{2}}{4 A^{2}} = \frac{4}{1}

💡 Teacher's Secret Hint

Ensure to square the sum/difference of amplitudes, not just the amplitudes themselves.

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Physics Question 41 – JEE-MAIN 2025

Clue 1 of 3: Understand the relationship between intensity and amplitude

🧠 Full Solution Path