| ⇦ | 
| ⇨ |
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: - The ionisation potential of hydrogen-atom is -13.6 eV. An electron in the groundstate
- A man weighing 60 kg climbs a staircase carrying a 20 kg load on his hand
- An electron of mass Mₑ, initially at rest, moves through a certain distance in a uniform
- The power factor of an AC circuit having resistance R and inductance L connected
- Three identical spheres, each of mass 3 kg are placed touching each other
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The ionisation potential of hydrogen-atom is -13.6 eV. An electron in the groundstate
- A man weighing 60 kg climbs a staircase carrying a 20 kg load on his hand
- An electron of mass Mₑ, initially at rest, moves through a certain distance in a uniform
- The power factor of an AC circuit having resistance R and inductance L connected
- Three identical spheres, each of mass 3 kg are placed touching each other
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