A normal eye has its retina positioned 2 cm behind the eye-lens. Determine the power of the eye-lens when the eye is fully relaxed.

Correct answer: B. 50 D When the eye is fully relaxed, it focuses parallel rays from an object at infinity onto the retina. Object distance, u = -∞ Image distance, v = 2 cm = 0.02 m Using the lens formula: (1)/(f) = (1)/(v) - (1)/(u) (1)/(f) = (1)/(0.02) - (1)/(-∞) (1)/(f) = 50 m^ -1 The power of the eye-lens is given by: P = (1)/(f) = 50 D

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Optical Instruments

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Optical Instruments

A normal eye has its retina positioned 2 cm behind the eye-lens....

A normal eye has its retina positioned

2 \text{ cm}

behind the eye-lens. Determine the power of the eye-lens when the eye is fully relaxed.

Options

A $46 \text{ D}$
B $50 \text{ D}$
C $54 \text{ D}$
D $40 \text{ D}$

Correct answer

B. $50 \text{ D}$

Step-by-step solution

When the eye is fully relaxed, it focuses parallel rays from an object at infinity onto the retina.

Object distance,

u = -\infty

Image distance,

v = 2 \text{ cm} = 0.02 \text{ m}

Using the lens formula:

\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}

\dfrac{1}{f} = \dfrac{1}{0.02} - \dfrac{1}{-\infty}

\dfrac{1}{f} = 50 \text{ m}^{-1}

The power of the eye-lens is given by:

P = \dfrac{1}{f} = 50 \text{ D}

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