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. A slab consists of two parallel layers of two different materials of same thickness
  2. α-particles, deuterons and protons of same energy are moving in a perpendicular
  3. A geostationary satellite is orbiting the earth at a height of 5R above that surface
  4. A device which converts electrical energy into mechanical energy is
  5. A round uniform body of radius R, mass M and moment of inertia I rolls down

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Your email address will not be published.


*