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The Human Eye and Optical Phenomena - The Human Eye and the Colourful World

Grade 10CBSE

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

🔑Concepts

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The Human Eye: A natural optical instrument. Key parts include the Cornea (outer protective layer), Iris (controls pupil size), Pupil (regulates light), Crystalline Lens (converging lens), and Retina (screen where images form).

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Power of Accommodation: The ability of the eye lens to adjust its focal length to see both nearby and distant objects clearly. The least distance of distinct vision (Near Point) for a normal eye is 25 cm25\text{ cm}.

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Myopia (Near-sightedness): A defect where a person can see nearby objects clearly but cannot see distant objects. The image is formed in front of the retina. It is corrected using a concave lens of suitable focal length.

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Hypermetropia (Far-sightedness): A defect where a person can see distant objects clearly but cannot see nearby objects. The image is formed behind the retina. It is corrected using a convex lens.

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Presbyopia: Gradual weakening of ciliary muscles and diminishing flexibility of the eye lens due to aging. It is often corrected using bifocal lenses.

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Refraction through a Prism: When light passes through a glass prism, it bends. The angle between the incident ray and the emergent ray is called the Angle of Deviation (δ\delta).

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Dispersion: The splitting of white light into its component colors (VIBGYOR). Red light bends the least (longer wavelength), while violet light bends the most (shorter wavelength).

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Atmospheric Refraction: Phenomena like the twinkling of stars, advanced sunrise (by 22 minutes), and delayed sunset (by 22 minutes) are caused by the refraction of light through layers of the atmosphere with varying refractive indices.

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Tyndall Effect: The scattering of light by colloidal particles. The color of scattered light depends on the size of the particles: very fine particles scatter blue light, while larger particles scatter longer wavelengths (red).

📐Formulae

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

P=1f(m)P = \frac{1}{f(m)}

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

Pnet=P1+P2+P3+…P_{net} = P_1 + P_2 + P_3 + \dots

💡Examples

Problem 1:

The far point of a myopic person is 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 myopic eye, to see distant objects, the object distance is u=−∞u = -\infty. The image must be formed at the far point, so v=−80 cm=−0.8 mv = -80\text{ cm} = -0.8\text{ m}. Using the lens formula: 1f=1v−1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u} 1f=1−0.8−1−∞\frac{1}{f} = \frac{1}{-0.8} - \frac{1}{-\infty} 1f=−1.25\frac{1}{f} = -1.25 f=−0.8 mf = -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 a concave (diverging) lens is required. The power needed is −1.25 Dioptres-1.25\text{ Dioptres}.

Problem 2:

A person needs a lens of power −5.5 dioptres-5.5\text{ dioptres} for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre+1.5\text{ dioptre}. What is the focal length of the lens required for correcting distant vision?

Solution:

For distant vision: Power P=−5.5 DP = -5.5\text{ D}. We know that f=1Pf = \frac{1}{P} f=1−5.5 mf = \frac{1}{-5.5}\text{ m} f≈−0.1818 m=−18.18 cmf \approx -0.1818\text{ m} = -18.18\text{ cm}

Explanation:

Since the power is negative, the lens used for distant vision is a concave lens with a focal length of approximately −18.2 cm-18.2\text{ cm}.