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JEE Main 2021 (1 Sep, Shift 2) · Superposition, Coherence & Intensity

Superposition, Coherence & Intensity question from JEE Main 2021 (1 Sep, Shift 2)

Q79JEE Main 2021 · 1 SEP · SHIFT 2Medium

Superposition, Coherence & Intensity

The width of one of the two slits in a Young's double slit experiment is three times the other slit. If the amplitude of the light coming from a slit is proportional to the slit-width, the ratio of minimum to maximum intensity in the interference pattern is x: 4 where x is ____________.

Numerical answer type. Work it out, then open the answer.

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Answer: 1

  1. Amplitude ∝\propto slit width, so a1:a2=1:3a_1 : a_2 = 1 : 3.
  2. IminImax=(a2−a1)2(a2+a1)2=(3−1)2(3+1)2=416=14\dfrac{I_{min}}{I_{max}} = \dfrac{(a_2 - a_1)^2}{(a_2 + a_1)^2} = \dfrac{(3-1)^2}{(3+1)^2} = \dfrac{4}{16} = \dfrac{1}{4}
  3. Comparing with x:4x : 4 gives x=1x = 1.
  4. So the answer is 11.
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