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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
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Topics: Ray Optics
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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The amplitude of S.H.M. y=2 (sin 5πt+√2 cos 5πt) is
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- A particle is moving such that its position coordinates (x,y) are (2 m,3 m) at time
- A neutron moving with velocity v collides with a stationary particle
- 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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