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

In Youngs Experiment The Ratio Of Maximum To Minimum Intensities Physics Question

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. If the band gap between valence band and conduction band in a material is 5.0 eV,
  2. An electric lamp is connected to 220 V, 50 Hz supply. Then the peak voltge is
  3. A heavy stone hanging from a massless string of length 15 m is projected
  4. The approximate depth of an ocean is 2700 m. The compressibility of water is
  5. 10 mA current can pass through a galvanometer of resistance 25 ohm.

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.

NEETLab Mobile App

Share this page with your friends

Be the first to comment

Leave a Reply

Your email address will not be published.


*