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The Human Eye and the Colourful World - Functioning of a lens in human eye

Grade 10CBSE

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

🔑Concepts

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The human eye contains a crystalline lens which is a convex lens made of a transparent, flexible, jelly-like material.

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Image Formation: The eye lens forms a real and inverted image of objects on the retina, which acts as a light-sensitive screen.

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Ciliary Muscles: These muscles control the curvature and focal length (ff) of the eye lens. When the muscles relax, the lens becomes thin and its focal length increases to see distant objects clearly. When the muscles contract, the lens becomes thick and its focal length decreases to see nearby objects.

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Power of Accommodation: This is the ability of the eye lens to adjust its focal length to see both nearby and distant objects clearly. The minimum distance at which objects can be seen most distinctly without strain is the Near Point (Least Distance of Distinct Vision), which is 25 cm25\text{ cm} for a normal eye.

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The Far Point of a normal human eye is at infinity (∞\infty).

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The light-sensitive cells on the retina (rods and cones) get activated upon illumination and generate electrical signals, which are sent to the brain via the optic nerve.

📐Formulae

Lens Formula: 1f=1v−1u\text{Lens Formula: } \frac{1}{f} = \frac{1}{v} - \frac{1}{u}

Power of a Lens (P): P=1f (in metres)\text{Power of a Lens (P): } P = \frac{1}{f \text{ (in metres)}}

Magnification (m): m=vu\text{Magnification (m): } m = \frac{v}{u}

💡Examples

Problem 1:

A person with a normal eye is looking at an object placed at the near point (25 cm25\text{ cm}). If the distance between the eye lens and the retina is 2.5 cm2.5\text{ cm}, calculate the focal length of the eye lens in this state.

Solution:

Given: Object distance u=−25 cmu = -25\text{ cm} (sign convention), Image distance v=+2.5 cmv = +2.5\text{ cm} (distance to retina). Using the lens formula: 1f=1v−1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u} 1f=12.5−1−25\frac{1}{f} = \frac{1}{2.5} - \frac{1}{-25} 1f=1025+125=1125\frac{1}{f} = \frac{10}{25} + \frac{1}{25} = \frac{11}{25} f=2511≈2.27 cmf = \frac{25}{11} \approx 2.27\text{ cm}

Explanation:

The eye lens adjusts its focal length to 2.27 cm2.27\text{ cm} to focus the object placed at the near point exactly onto the retina.

Problem 2:

Explain the state of the eye lens when focusing on an object at infinity (∞\infty).

Solution:

For an object at infinity, u=−∞u = -\infty. Using the lens formula: 1f=1v−1−∞\frac{1}{f} = \frac{1}{v} - \frac{1}{-\infty} Since 1∞=0\frac{1}{\infty} = 0, we get 1f=1v  ⟹  f=v\frac{1}{f} = \frac{1}{v} \implies f = v

Explanation:

In this case, the focal length of the eye lens becomes equal to the distance between the lens and the retina (f≈2.5 cmf \approx 2.5\text{ cm}). The ciliary muscles are in their most relaxed state, making the lens thin.