| ⇦ | 
| ⇨ |
The intensity at the maximum in a Young’s double slit experiment is I₀. Distance between two slits is d=5λ, where λ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance, D= 10 d?
Options
(a) I₀/4
(b) 3/4 I₀
(c) I₀/2
(d) I₀
Correct Answer:
I₀/2
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - The velocity of a particle performing simple harmonic motion, when it passes
- Four cells, each of emf E and internal resistance r, are connected in series across
- The horizontal range of a projectile is 4√3 times the maximum height achieved
- When germanium is doped with phosphorus what type of semiconductor is produced?
- A battery of emf 10 V and internal resistance 3 ohm is connected to a resistor.
Topics: Wave Optics
(101)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The velocity of a particle performing simple harmonic motion, when it passes
- Four cells, each of emf E and internal resistance r, are connected in series across
- The horizontal range of a projectile is 4√3 times the maximum height achieved
- When germanium is doped with phosphorus what type of semiconductor is produced?
- A battery of emf 10 V and internal resistance 3 ohm is connected to a resistor.
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
Leave a Reply