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The Human Eye and the Colourful World - Defects of vision and their corrections

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Power of Accommodation is the ability of the eye lens to adjust its focal length (ff) so as to clearly focus images of objects at various distances on the retina.

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Myopia (Near-sightedness): A person can see nearby objects clearly but cannot see distant objects distinctly. The image is formed in front of the retina. This is corrected using a concave lens of suitable power (P<0P < 0).

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Hypermetropia (Far-sightedness): A person can see distant objects clearly but cannot see nearby objects distinctly. The image is formed behind the retina. This is corrected using a convex lens of suitable power (P>0P > 0).

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Presbyopia: The power of accommodation of the eye usually decreases with aging. The near point gradually recedes. It is often corrected using bifocal lenses, where the upper portion is a concave lens and the lower portion is a convex lens.

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The Far Point for a normal eye is infinity (∞\infty), and the Near Point (Least Distance of Distinct Vision) is 25 cm25\text{ cm}.

📐Formulae

1f=1v−1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

P=1f (in meters)P = \frac{1}{f \text{ (in meters)}}

P=100f (in cm)P = \frac{100}{f \text{ (in cm)}}

Power (Dioptre, D)=P1+P2+...+Pn\text{Power (Dioptre, D)} = P_1 + P_2 + ... + P_n

💡Examples

Problem 1:

A myopic person has a far point of 80 cm80\text{ cm} in front of the eye. What is the nature and power of the lens required to enable him to see very distant objects clearly?

Solution:

For a distant object, u=−∞u = -\infty. The image should be formed at the far point, so v=−80 cmv = -80\text{ cm}. Using lens formula: 1f=1v−1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u} 1f=1−80−1−∞\frac{1}{f} = \frac{1}{-80} - \frac{1}{-\infty} 1f=−180  ⟹  f=−80 cm=−0.8 m\frac{1}{f} = -\frac{1}{80} \implies f = -80\text{ cm} = -0.8\text{ m}. Power P=1f=1−0.8=−1.25 DP = \frac{1}{f} = \frac{1}{-0.8} = -1.25\text{ D}.

Explanation:

The negative sign indicates that the lens is a concave (diverging) lens, which is required to shift the image back onto the retina.

Problem 2:

The near point of a hypermetropic eye is 1 m1\text{ m}. What is the power of the lens required to correct this defect? Assume the near point of the normal eye is 25 cm25\text{ cm}.

Solution:

The object is placed at the normal near point, u=−25 cmu = -25\text{ cm}. The lens must form an image at the person's near point, v=−1 m=−100 cmv = -1\text{ m} = -100\text{ cm}. Using lens formula: 1f=1−100−1−25\frac{1}{f} = \frac{1}{-100} - \frac{1}{-25} 1f=−1100+4100=3100\frac{1}{f} = -\frac{1}{100} + \frac{4}{100} = \frac{3}{100} f=1003 cm=13 mf = \frac{100}{3}\text{ cm} = \frac{1}{3}\text{ m}. Power P=1f=+3.0 DP = \frac{1}{f} = +3.0\text{ D}.

Explanation:

The positive sign indicates a convex (converging) lens, which helps the eye focus light from nearby objects onto the retina.

Defects of vision and their corrections Revision - Class 10 Science CBSE