Physics Q41 – JEE-MAIN 2025

Two plane polarized light waves combine at a certain point whose electric field components are

E_{1} = E_{0} \sin ω t

E_{2} = E_{0} \sin (ω t + \frac{π}{3})

Find the amplitude of the resultant wave.

Clue 1 of 3: Superposition Principle

When two waves combine, their resultant amplitude depends on their individual amplitudes and the phase difference between them.

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Ninja Strategy•Range Estimation

Estimate the possible range of the resultant amplitude based on the individual amplitudes and phase difference to eliminate options that are too large or too small.

🧠 Full Solution Path

Step 1: Identify Wave Parameters✦ Active

The given electric field components are $E_{1} = E_{0} \sin ω t$ and $E_{2} = E_{0} \sin (ω t + \frac{π}{3})$ . From these, we identify the individual amplitudes as $A_{1} = E_{0}$ and $A_{2} = E_{0}$ . The phase difference between the two waves is $ϕ = (ω t + \frac{π}{3}) - ω t = \frac{π}{3}$ .

Step 2: Apply Resultant Amplitude Formula○ Expand

The amplitude of the resultant wave $A$ is given by the formula:

A = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}

Substitute the identified values:

A = \sqrt{E_{0}^{2} + E_{0}^{2} + 2 E_{0} E_{0} \cos (\frac{π}{3})}

Step 3: Calculate the Resultant Amplitude○ Expand

Knowing that $\cos (\frac{π}{3}) = \frac{1}{2}$ , substitute this into the equation:

A = \sqrt{2 E_{0}^{2} + 2 E_{0}^{2} (\frac{1}{2})} = \sqrt{2 E_{0}^{2} + E_{0}^{2}} = \sqrt{3 E_{0}^{2}} = E_{0} \sqrt{3}

Using the approximation $\sqrt{3} \approx 1.732$ , the resultant amplitude is $A \approx 1.732 E_{0}$ . Comparing with the options, $1.7 E_{0}$ is the closest value.

💡 Teacher's Secret Hint

Remember to use the correct trigonometric value for the phase difference.

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Physics Question 41 – JEE-MAIN 2025

Clue 1 of 3: Superposition Principle

🧠 Full Solution Path