⇦ | ⇨ |
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 energy of groundstate (n=1) of hydrogen level is -13.6 eV. The ionistation
- A particle moves along a circle of radius 20/π m with constant tangential acceleration
- If the feedback voltage is increased in a negative feedback amplifier, then
- In Young’s double silt experiment if the distance between the two slits
- If the length of a stretched string is shortened by 4% and the tension is increased
Topics: Wave Optics
(101)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The energy of groundstate (n=1) of hydrogen level is -13.6 eV. The ionistation
- A particle moves along a circle of radius 20/π m with constant tangential acceleration
- If the feedback voltage is increased in a negative feedback amplifier, then
- In Young’s double silt experiment if the distance between the two slits
- If the length of a stretched string is shortened by 4% and the tension is increased
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