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NVAEasyJEE 2025Interference (Young's Experiment)

JEE Physics 2025 Question with Solution

Two coherent monochromatic light beams of intensities 4I4I and 9I9I are superimposed. The difference between the maximum and minimum intensities in the resulting interference pattern is xIxI. The value of xx is _____.

Answer

Correct answer:24

Step-by-step solution

Standard Method

Given: Two coherent monochromatic light beams have intensities I1=4II_1 = 4I and I2=9II_2 = 9I.

Find: The value of xx if the difference between maximum and minimum intensities is xIxI.

For interference of two coherent beams,

Imax⁡=(I1+I2)2I_{\max} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2

and

Imin⁡=(I1−I2)2I_{\min} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^2

Now,

I1=4I=2I\sqrt{I_1} = \sqrt{4I} = 2\sqrt{I} I2=9I=3I\sqrt{I_2} = \sqrt{9I} = 3\sqrt{I}

So,

Imax⁡=(2I+3I)2=(5I)2=25II_{\max} = \left(2\sqrt{I} + 3\sqrt{I}\right)^2 = \left(5\sqrt{I}\right)^2 = 25I

And,

Imin⁡=(3I−2I)2=(I)2=II_{\min} = \left(3\sqrt{I} - 2\sqrt{I}\right)^2 = \left(\sqrt{I}\right)^2 = I

Therefore, the difference is

Imax⁡−Imin⁡=25I−I=24II_{\max} - I_{\min} = 25I - I = 24I

Comparing with xIxI,

x=24x = 24

Therefore, the value of xx is 2424.

Direct Identity

Given: I1=4II_1 = 4I and I2=9II_2 = 9I.

Find: The value of xx in Imax⁡−Imin⁡=xII_{\max} - I_{\min} = xI.

Use the identity

Imax⁡−Imin⁡=(I1+I2)2−(I1−I2)2=4I1I2I_{\max} - I_{\min} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2 - \left(\sqrt{I_1} - \sqrt{I_2}\right)^2 = 4\sqrt{I_1 I_2}

Substitute the values:

Imax⁡−Imin⁡=4(4I)(9I)=436I2=24II_{\max} - I_{\min} = 4\sqrt{(4I)(9I)} = 4\sqrt{36I^2} = 24I

Hence,

x=24x = 24

This works because the difference of the constructive and destructive intensity expressions reduces immediately by the algebraic identity (a+b)2−(a−b)2=4ab(a+b)^2 - (a-b)^2 = 4ab.

Common mistakes

  • Using Imax⁡=I1+I2I_{\max} = I_1 + I_2 and Imin⁡=I1−I2I_{\min} = I_1 - I_2 directly is incorrect because interference depends on amplitudes, not just algebraic addition or subtraction of intensities. First convert intensities to amplitude form through square roots.

  • Writing Imin⁡=(2I−3I)=−II_{\min} = (2\sqrt{I} - 3\sqrt{I}) = -\sqrt{I} as the final intensity is wrong because intensity must be the square of the resultant amplitude. The correct step is Imin⁡=(−I)2=II_{\min} = (-\sqrt{I})^2 = I.

  • Comparing 24I24I with xIxI and keeping the answer as 24I24I is incorrect. In a numerical value answer, xx is only the coefficient, so the answer is 2424.

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