In Young’s experiment, the ratio of maximum to minimum intensities of the fringe

⇦

⇨

In Young’s experiment, the ratio of maximum to minimum intensities of the fringe system is 4:1. The amplitudes of the coherent sources are in the ratio

Options

(a) (4:1)
(b) (3:1)
(c) (2:1)
(d) (1:1)

Correct Answer:

(3:1)

Explanation:

I(max) / I(min) = (ɑ₁ + ɑ₂)² / (ɑ₁ – ɑ₂)²

⇒ 4 / 1 = (ɑ₁ + ɑ₂)² / (ɑ₁ – ɑ₂)²

⇒ (ɑ₁ + ɑ₂) / (ɑ₁ – ɑ₂) = 2 / 1

⇒ ɑ₁ + ɑ₂ = 2 (ɑ₁ – ɑ₂)

⇒ 3ɑ₂ = ɑ₁

⇒ ɑ₁ / ɑ₂ = 3 / 1 = 3 : 1

Related Questions:

1. Two bodies of mass 1 kg and 3 kg have positive vectors i + 2j + k and -3i -2j + k
2. The following four wires are made of the same material. Which of these will have
3. The work function of sodium is 2.3 eV. The threshold wavelength of sodium will be
4. The speed of light in media M₁ and M₂ are 1.5×10⁸ ms⁻¹ and 2×10⁸ ms⁻¹ respectively.
5. The equivalent resistance of two resistors connected in series is 6 Ω

Topics: Wave Optics (101)
Subject: Physics (2479)

---

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

---

Share this page with your friends