krit.club logo

The Human Eye and Optical Phenomena - THE HUMAN EYE

Grade 10CBSE

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

🔑Concepts

•

The human eye is like a camera. Its lens system forms an image on a light-sensitive screen called the retina. The eyeball is approximately spherical in shape with a diameter of about 2.3 cm2.3\text{ cm}.

•

Cornea: A thin membrane through which light enters. Most of the refraction for the light rays entering the eye occurs at the outer surface of the cornea.

•

Iris and Pupil: The iris is a dark muscular diaphragm that controls the size of the pupil. The pupil regulates and controls the amount of light entering the eye.

•

Eye Lens: A convex lens made of transparent, flexible, jelly-like material. It forms an inverted real image of the object on the retina.

•

Ciliary Muscles: These muscles modify the curvature of the eye lens to adjust its focal length, allowing us to see both nearby and distant objects clearly.

•

Power of Accommodation: The ability of the eye lens to adjust its focal length is called accommodation. When muscles are relaxed, ff increases (for distant objects); when muscles contract, ff decreases (for nearby objects).

•

Near Point and Far Point: The minimum distance at which objects can be seen most distinctly without strain is the least distance of distinct vision, D=25 cmD = 25\text{ cm}. The far point for a normal eye is infinity (∞\infty).

•

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. It is corrected using a concave lens.

•

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

•

Presbyopia: A defect occurring due to aging where the power of accommodation decreases. It often requires bifocal lenses (upper part concave, lower part convex).

📐Formulae

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

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

For Myopia correction: f=−d (where d is the far point of the defective eye)\text{For Myopia correction: } f = -d \text{ (where } d \text{ is the far point of the defective eye)}

For Hypermetropia correction: 1f=1v−1u where u=−25 cm and v=−(near point)\text{For Hypermetropia correction: } \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \text{ where } u = -25\text{ cm and } v = -(\text{near point})

💡Examples

Problem 1:

A person with a myopic eye 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 distinctly?

Solution:

For a myopic eye, to see distant objects, the object distance u=−∞u = -\infty. The image should 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+0\frac{1}{f} = -1.25 + 0 f=−0.8 mf = -0.8\text{ m} P=1f=1−0.8=−1.25 DP = \frac{1}{f} = \frac{1}{-0.8} = -1.25\text{ D}

Explanation:

Since the power is negative, the lens required is a concave (diverging) lens with a power of −1.25 Dioptres-1.25\text{ Dioptres}.

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 that the near point of the normal eye is 25 cm25\text{ cm}.

Solution:

Here, the object distance u=−25 cm=−0.25 mu = -25\text{ cm} = -0.25\text{ m} (normal near point). The image must be formed at the defective eye's near point, v=−1 mv = -1\text{ m}. Using the lens formula: 1f=1v−1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u} 1f=1−1−1−0.25\frac{1}{f} = \frac{1}{-1} - \frac{1}{-0.25} 1f=−1+4\frac{1}{f} = -1 + 4 1f=3\frac{1}{f} = 3 P=+3.0 DP = +3.0\text{ D}

Explanation:

A positive power indicates a convex (converging) lens is needed. The person requires a convex lens of +3.0 D+3.0\text{ D} to correct the hypermetropia.