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) 2 (b) 4 (c) 18 (d) 9

Answer: (b). ImaxImin = ((a₁+a₂)/(a₂-a₁))² with a₁: a₂ = 1: 3 = ((1+3)/(3-1))² = 16/4 = 4 So the answer is (b) 4.

In an interference experiment the ratio of amplitudes of coherent… | JEE Main 2019 (8 Apr, Shift 1)

Q89JEE Main 2019 · 8 APR · SHIFT 1Easy

Superposition, Coherence & Intensity

In an interference experiment the ratio of amplitudes of coherent waves is $\frac{a_1}{a_2} = \frac{1}{3}$ . The ratio of maximum and minimum intensities of fringes will be:

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Answer: (b)

$\frac{I_{max}}{I_{min}} = \left(\frac{a_1+a_2}{a_2-a_1}\right)^2$ with $a_1 : a_2 = 1 : 3$
$= \left(\frac{1+3}{3-1}\right)^2 = \frac{16}{4} = 4$
So the answer is (b) 4.

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