| ⇦ | 
| ⇨ |
The diameter of the eye ball of a normal eye is about 2.5 cm. The power of the eye lens varies from
Options
(a) 9 D to 8 D
(b) 40 D to 32 D
(c) 44 D to 40 D
(d) None of these
Correct Answer:
44 D to 40 D
Explanation:
An eye see distant objects with full relaxation.
So, [1 / (2.5 × 10⁻²)] – [1 / -∞] = 1 / f or, P = 1 / f = 1 / 2.5 × 10⁻² = 40 D
An eye see an object at 25 cm with strain [1 / (2.5 × 10⁻²)] – [1 / 2.5 × 10⁻²] = 1 / f
.·. P = 1 / f = 40 + 4 = 44 D
Related Questions: - A short magnet of magnetic moment M, is placed on a straight line. The ratio
- Two identical flutes produce fundamental notes of frequency 300 Hz at 27° C
- Which of the following figures represent the variation of particle momentum
- A Carnot engine, having an efficiency of η=1/10 as heat engine, is used
- The wheel of a car is rotating at the rate of 1200 rev/min. On pressing
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A short magnet of magnetic moment M, is placed on a straight line. The ratio
- Two identical flutes produce fundamental notes of frequency 300 Hz at 27° C
- Which of the following figures represent the variation of particle momentum
- A Carnot engine, having an efficiency of η=1/10 as heat engine, is used
- The wheel of a car is rotating at the rate of 1200 rev/min. On pressing
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