The two light beams having intensities I and 9I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π/2 at point P and π at point Q. Then the difference between the resultant intensities at P and Q will be :

  1. 2I
  2. 6I
  3. 5I
  4. 7I

Answer (Detailed Solution Below)

Option 2 : 6I
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JEE Main 04 April 2024 Shift 1
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Detailed Solution

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CONCEPT:

  • The intensity is directly proportional to the square of the amplitude and it is written as;

            I ∝ A2

           Here we have "I" as the intensity and A as the amplitude.

  • The resultant intensity is written as:

           I = \({I_1+I_2+2\sqrt {I_1I_2}cos ϕ}\)

          Here we have I1, I2 are the intensities of two beams, and ϕ is the phase difference.

CALCULATION:

Given:

The intensity of the first beam, I1 = I

The intensity of the second beam, I2 = 9I

The phase difference at point P = π/2 

and the phase difference at point Q = π 

Now, the resultant intensity at point P with a phase difference π/2,

Ip = \({I+9I+2\sqrt {I\times 9I}cos \frac{π}{2}}\)

Ip = 10I

Ip = 10I

Now, the resultant intensity at point P with a phase difference π, 

Iq = \({I+9I+2\sqrt {I\times 9I}cos{π}}\)

Iq = 10I - 6I

Iq = 4I

Now, the difference in the intensity of beam P and Q we have;

I = Ip - Iq

⇒ I = 10I - 4I

⇒ I = 6I

Hence option 2) is the correct answer.

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