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In Young’s double-slit experiment, the two coherent sources have different intensities. If the ratio of the maximum intensity to the minimum intensity in the interference pattern produced is 25:1, - Physics

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Question

In Young’s double-slit experiment, the two coherent sources have different intensities. If the ratio of the maximum intensity to the minimum intensity in the interference pattern produced is 25:1, what is the ratio of the intensities of the two sources?

Options

  • 5:1

  • 25:1

  • 3:2

  • 9:4

MCQ
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Solution

9:4

Explanation:

Given: The ratio of maximum intensity to minimum intensity is `I_"max"/I_"min" = 25/1`

Apply the formula to calculate the ratio `I_"max"/I_"min"`

`I_"max"/I_"min" = (a_1 + a_2)^2/(a_1 - a_2)^2`

Where, `a_1^2/a_2^2 = I_1/I_2`

Substitute `I_"max"/I_"min" = 25/1` for `I_"max"/I_"min" = 25/1` in the above relation.

`25/1 = (a_1 + a_2)^2/(a_1 - a_2)^2`

`5/1 = (a_1 + a_2)/(a_1 - a_2)`

⇒ a1 = 3; a2 = 2

Substitute 3 for a1 and 2 for a2 in the relation `a_1^2/a_2^2 = I_1/I_2`

`I_1/I_2 = 3^2/2^2`

`I_1/I_2 = 9/4`

I1 : I2 :: 9 : 4

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Chapter 7: Wave Optics - Exercises [Page 184]

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Balbharati Physics [English] Standard 12 Maharashtra State Board
Chapter 7 Wave Optics
Exercises | Q 1.4 | Page 184

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