| ⇦ | 
| ⇨ |
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: - An object moving with a speed of 6.25 m/s, is deaccelerated at a rate given by
- The rest mass of a body is m1. It moves with a velocity of 0.6c, then its relativistic
- The binding energy per nucleon of ₃⁷Li and ₂⁴He nuclei are 5.60 MeV and 7.06MeV
- The frequencies of X-rays,γ-rays and ultraviolet rays are respectively p,q and r
- The S.I. unit of electric flux is
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- An object moving with a speed of 6.25 m/s, is deaccelerated at a rate given by
- The rest mass of a body is m1. It moves with a velocity of 0.6c, then its relativistic
- The binding energy per nucleon of ₃⁷Li and ₂⁴He nuclei are 5.60 MeV and 7.06MeV
- The frequencies of X-rays,γ-rays and ultraviolet rays are respectively p,q and r
- The S.I. unit of electric flux is
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