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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 binding energy per nucleon of ₃⁷Li and ₂⁴He nuclei are 5.60 MeV and 7.06MeV
- A solid cylinder of mass 50kg and radius 0.5 m is, free to rotate about the horizontal axis
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Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The binding energy per nucleon of ₃⁷Li and ₂⁴He nuclei are 5.60 MeV and 7.06MeV
- A solid cylinder of mass 50kg and radius 0.5 m is, free to rotate about the horizontal axis
- A body undergoes no change in volume. Poisson’s ratio is
- A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. The work
- A bomb of mass 16 kg at rest explodes into two pieces of masses of 4 kg and 12 kg.
Topics: Ray Optics (94)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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