Question

In an interference experiment the ratio of amplitudes of coherent waves is a₁/a₂ = 1/3 . The ratio of maximum and minimum intensities of fringes will be:
(A) 9 (B) 4
(C) 18 (D) 2

Answer 1

Explanation:

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Answer 2

The ratio of maximum and minimum intensity of fringes = 4.

Option (B) is correct.

Explanation:

Given :

The amplitude of one wave $(a_{1}) =$ 1.

The amplitude of another wave $(a_{2}) =$ 3.

The waves is coherent so in coherent wave frequency of wave is constant at any time.

From the equation,

⇒ $I _{max} = (\sqrt{I_{1} }+\sqrt{I_{2} } )^{2}$

⇒  $I _{min} = (\sqrt{I_{1} }-\sqrt{I_{2} } )^{2}$

We know that intensity is proportional to the square of the amplitude.

So we can write for maximum intensity,

  $I _{max} = (a_{1} }+a_{2} } )^{2}$

  $I_{max} = (4)^{2} = 16$

For minimum intensity,

  $I _{min} = (a_{1} }-a_{2} } )^{2}$

  $I_{min} = (-2)^{2} = 4$

Now divide maximum intensity to the minimum intensity.

  $\frac{I_{max} }{I_{min} } = \frac{16}{4} = 4$

Thus, the ratio of maximum and minimum intensity of fringes = 4.