| ⇦ | 
| ⇨ |
If μᵥ=1.5230 and μʀ=15.145, then dispersive power of crown glass is
Options
(a) 0.0164
(b) 0.00701
(c) 0.0132
(d) 0.032
Correct Answer:
0.0164
Explanation:
Here μᵥ=1.5230and μʀ=1.5145
Therefore, mean refractive index, μ =(1.523+1.5145) /2 = 1.5187
Now, dispersive power, ω is given by,
ω =(μᵥ-μʀ)/(μ-1) =(1.5230 – 1.5145)/(1.5187 – 1) =0.0164
Related Questions: - Which of the following is an example of ideal black body?
- What is the amount of energy released by deuterium and tritium fusion?
- The dual nature of light is exhibited by
- An unpolarised beam of intensity I₀ falls on a polaroid. The intensity
- An air core solenoid has 1000 turns and is one meter long. Its cross sectional area
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which of the following is an example of ideal black body?
- What is the amount of energy released by deuterium and tritium fusion?
- The dual nature of light is exhibited by
- An unpolarised beam of intensity I₀ falls on a polaroid. The intensity
- An air core solenoid has 1000 turns and is one meter long. Its cross sectional area
Topics: Ray Optics (94)
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