⇦ | ![]() | ⇨ |
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:
- When a bullet of mass 10 g and speed 100 ms⁻¹ penetrates up to distance 1 cm
- A pellet of mass 1 g is moving with an angular velocity of 1 rad/s along a circle
- If potential(in volts) in a region is expressed as V(x,y,z) = 6xy – y + 2yz,
- A motor-cyclist drives a motor cycle in a vertical circle. His minimum velocity
- Three capacitors each of capacitance C and of breakdown voltage V are joined in series
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