Human Eye and Colorful World Class-10 Notes & Numerical

 

Human Eye and Colorful World Class-10 Notes & Numerical

Introduction:

                            The chapter 'The Human Eye and the Colourful World' deals with the structure and working of the human eye, defects of vision, and various optical phenomena in nature. It connects physics with biology and explains how we perceive the world around us in colour.

Structure of the Human Eye:

                    The human eye is a spherical organ (about 2.5 cm in diameter) that acts as a natural optical device. It works like a camera – it receives light, focuses it, and forms an image on the retina.

Main Parts of the Human Eye:

1) Cornea

• Transparent, curved front surface of the eye. 
• Allows light to enter the eye.
• Provides most of the refraction (bending of light rays). 
• Helps in focusing light on the retina.

2) Iris

• Colored circular part of the eye (brown, black, blue, etc.). 
• Controls the size of the pupil using tiny muscles.
• Regulates the amount of light entering the eye.

3) Pupil

• Black circular opening in the center of the iris. 
• Expands in dim light (dilates) and contracts in bright light (constricts). 
• Controls the entry of light into the eye.

4) Lens

• Transparent, convex, and flexible structure. 
• Focuses light onto the retina by changing its curvature. 
• Controlled by ciliary muscles which adjust focal length for near and distant vision                     (Process : accommodation) 

5) Ciliary Muscles

• Ring-shaped muscles attached to the lens. 
• Control the shape and focal length of the lens.
• Contract for near objects (lens become thick).
• Relax for distant objects (Lens become thin).

6) Aqueous Humour

• Clear watery fluid between cornea and lens. 
• Maintains pressure in the eye and provides nutrients to lens and cornea.

7) Vitreous Humour

• Transparent jelly-like substance between lens and retina. 
• Maintains shape of the eyeball and provides support.

8) Retina

• Innermost light-sensitive layer of the eye.
• Contains rods (dim light, black & white vision) and cones (bright light, color vision). 
• Converts light into electrical signals.

9) Optic Nerve

• Connects the eye to the brain 
• Transmits visual signals from retina to the brain for interpretation.

10) Sclera

• Tough, white outer covering of the eyeball. 
• Protects the inner parts of the eye 
• Also called the 'white of the eye'.

11) Choroid

• Middle layer between sclera and retina. 
• Rich in blood vessels, supplies oxygen and nutrients, 
• Prevents internal reflection.

12) Blind Spot

• Point on the retina where optic nerve leaves the eye. 
• No rods and cones present, hence no vision possible at this spot.

13) Yellow Spot (Macula/Fovea Centralis)

• Small central pit in the retina.
• Densely packed with cones, responsible for sharp central vision. 
• Best color and detail vision is perceived here.

Key Point: The image formed on the retina is real, inverted, and smaller. Our brain interprets it as upright.

Functioning of the Human Eye:

• The cornea and the eye lens refract light and form a real, inverted image of the object on the retina.
• The retina converts this image into electrical signals.
• These signals are sent to the brain via the optic nerve, and the brain interprets them as upright images.

Power of Accommodation:

Definition: The ability of the eye lens to adjust its focal length so that objects at different distances can be seen clearly is called the power of accommodation.

• Near point: The closest distance at which an object can be seen clearly (about 25 cm).
• Far point: The maximum distance up to which an object can be seen clearly (infinity).

Defects of Vision and their Correction:

1) Myopia (Nearsightedness): 

Definition:

Myopia is a defect of vision in which a person can see nearby objects clearly but distant objects appear blurred. It is one of the most common refractive errors of the human eye.

Causes of Myopia

Myopia occurs when light rays from a distant object focus in front of the retina instead of directly on it. This happens due to:

1. Elongation of Eyeball – The distance between the lens and the retina increases.

2. Excessive Curvature of Cornea/Lens – The cornea or lens becomes too curved, increasing the converging power of the eye lens.

Image Formation in Myopic Eye

- For distant objects: Image is formed in front of the retina → blurred vision.

- For nearby objects: Image is formed on the retina → clear vision.

Correction of Myopia

• Myopia is corrected using a concave lens (diverging lens) of suitable focal length.
• The concave lens diverges the parallel rays of distant objects before they enter the eye. This ensures that the image is formed on the retina instead of in front of it.

Key Points

- The near point of a normal eye = 25 cm.

- In myopia, the far point is closer than infinity (could be 2 m, 5 m, etc.).

- Concave lens power needed: P = 1/f = 1/D, where D = far point of defective eye (in meters).

Useful facts & formulas :

• Myopia (short-sightedness): a person can see nearby objects clearly but distant objects appear blurred. Far point (F') is < ∞ (finite).

• Correction: use a concave (diverging) lens so that rays from a distant object (at ∞) are focused at the eye's far point F'.

• For object at ∞, image distance v = −(far point) (virtual image on object side). Using thin lens formula: 1/f = 1/v − 1/u. For u = −∞, 1/u = 0, so 1/f = 1/v. Thus f = v. Power P = 1/f (in metres). For concave lens, f < 0 and P < 0.

Solved Numerical :

Problem 1: A myopic person has a far point of 1.0 m. What is the power of the concave lens required to correct the vision for distant objects?

Solution:

Given: Far point = 1.0 m. For object at infinity (u = −∞), image formed at v = −1.0 m (virtual image on object side).

Using 1/f = 1/v − 1/u and 1/u = 0 ⇒ 1/f = 1/(−1.0) = −1.0 ⇒ f = −1.0 m.

Power P = 1/f = 1/(−1.0) = −1.0 D.

Answer: −1.0 dioptre (concave lens).

Problem 2: A myopic person's far point is 50 cm. Find the focal length and power of the correcting lens.

Solution:

Given: Far point = 50 cm = 0.50 m ⇒ v = −0.50 m for object at infinity.

f = v = −0.50 m ⇒ P = 1/f = 1/(−0.50) = −2.0 D.

Answer: f = −0.50 m, P = −2.0 D.

Problem 3: A person can see clearly up to a distance of 2 m. Find the spectacles power needed to see distant objects clearly.

Solution:

Given far point = 2 m ⇒ v = −2.0 m. For u = −∞, f = −2.0 m.

Power P = 1/f = 1/(−2.0) = −0.5 D.

Answer: −0.5 D.

Problem 4: A student uses spectacles of power −1.5 D. What is his far point when wearing these spectacles? (Assume spectacles very near the eye.)

Solution:

Given P = −1.5 D ⇒ f = 1/P = −0.6667 m ≈ −66.67 cm. For distant object (u = −∞), image formed at v = f = −66.67 cm.

So far point with spectacles is ≈ 66.7 cm.

Answer: Far point ≈ 66.7 cm.

Problem 5: A myopic person's far point is 40 cm. What power of lens will enable him to see distant objects clearly?

Solution:

Far point = 40 cm = 0.40 m ⇒ f = −0.40 m ⇒ P = 1/(−0.40) = −2.5 D.

Answer: −2.5 D.

Problem 6: If a person’s far point is 25 cm, what spectacles power is needed to correct the vision?

Solution:

Far point = 25 cm = 0.25 m ⇒ f = −0.25 m ⇒ P = 1/(−0.25) = −4.0 D.

Answer: −4.0 D.

Problem 7: A myopic student has a far point of 1.2 m. He buys spectacles of power −0.75 D. Will these spectacles correct his distant vision?

Solution:

Given far point = 1.2 m ⇒ required correction for distant objects is P_required = 1/(−1.2) ≈ −0.833… D.

Given spectacles P = −0.75 D which is weaker (less negative) than required −0.833 D, so distant objects will still appear slightly blurred; spectacles partially correct but do not fully correct to infinity.

Answer: No — spectacles of −0.75 D under-correct; required ≈ −0.83 D.

Problem 8: A person can see clearly only up to 5 m. What is the power of the lens required to correct his vision for distant objects?

Solution:

Far point = 5 m ⇒ f = −5 m ⇒ P = 1/(−5) = −0.2 D.

Answer: −0.2 D.

Problem 9: A myopic eye has far point at 0.6 m. If the corrective lens is placed 2 cm from the eye, what power is needed? (Include vertex distance approx.)

Solution:

Far point of eye (without spectacles) = 0.6 m. Let the spectacles (placed 2 cm = 0.02 m from eye) form an image at the eye's far point. For object at infinity, the image distance from the lens should be v_lens measured from lens such that the image is at 0.6 m from the eye: distance from lens to image v_lens = −(0.6 − 0.02) = −0.58 m.

For u = −∞, 1/f = 1/v ⇒ f = v_lens = −0.58 m ⇒ P = 1/f ≈ 1/(−0.58) ≈ −1.724 D.

Answer: ≈ −1.72 D (taking vertex distance into account).

Problem 10: A boy can see objects clearly only up to a distance of 2 m. What type and power of lens does he need to see distant objects clearly?

Solution:

- Far point of defective eye = 2 m

- Normal far point = ∞

Required lens focal length (f) = 1/∞ - 1/2 = -1/2 m = –0.5 m

Power, P = 100/f (in cm) = 100 / –50 = –2 D

He needs a concave lens of power –2D.

Practice Problems (Try Yourself)

1) Far point = 75 cm. Find power of correcting lens.

2) A person’s far point is 1.5 m. What lens will correct distant vision?

3) A myopic eye has far point 30 cm. What is required power?

4) Spectacles of −3.0 D give a far point of what distance?

5) A myopic student with far point 90 cm wants to read at 25 cm. What lens would help (assume spectacles near the eye)?

6) If a lens of −2.0 D is placed 1.5 cm from the eye, what is the effective far point?

7) Far point = 200 cm. Find required power.

8) Person can see up to 10 m. What power is needed to correct distant objects?

9) A −0.5 D spectacle lens corrects the eye to infinity. What was the eye's far point?

10) A student with far point 60 cm uses spectacles of −1.8 D. Are these spectacles slightly over-correcting or under-correcting?

Answer Key (Practice)

1) f = −0.75 m ⇒ P = −1.33 D.

2) f = −1.5 m ⇒ P = −0.67 D.

3) f = −0.30 m ⇒ P = −3.33 D.

4) f = −0.333… m ⇒ far point ≈ 33.3 cm.

5) To see distant objects = P ≈ −1.11 D; for reading at 25 cm a different lens needed (complex).

6) f = −0.5 m approx ⇒ far point ≈ 50 cm from eye (account vertex distance properly).

7) f = −2.0 m ⇒ P = −0.5 D.

8) far point 10 m ⇒ P = −0.10 D.

9) P = −0.5 D ⇒ far point = 1/0.5 = 2.0 m.

10) Required for 60 cm is P = −1.67 D. Given −1.8 D is slightly more negative ⇒ slightly over-correcting.

2) Hypermetropia (Farsightedness):

Definition:

Hypermetropia (farsightedness) is a defect of vision in which a person can see distant objects clearly, but has difficulty in seeing nearby objects distinctly.

Causes of Hypermetropia

1. Eyeball is too short:
- The distance between the lens and retina is less than normal.
- Light rays from nearby objects focus behind the retina.

2. Eye lens is too thin or less converging:
- Insufficient curvature of the lens.
- Inability to bend light rays enough to focus them on the retina.

Image Formation in Hypermetropic Eye

- For near objects: Light rays converge behind the retina, image appears blurred.
- For distant objects: Light rays are nearly parallel and focused properly on the retina → clear vision.

Correction of Hypermetropia

• Convex lens (converging lens) is used in spectacles.
• The convex lens bends incoming light rays inward before entering the eye, so that they can focus on the retina properly.

Power of Correcting Lens

The focal length of the lens required is given by:
f = 1/P
Where:
- P = Power of lens (in dioptres)
- f = Focal length (in meters)

Useful facts & formulas About Hypermetropia :

• Hypermetropia (long-sightedness): near point D′ > 25 cm (normal near point D = 25 cm).

• Correction: convex (converging) lens so that an object at 25 cm forms a virtual image at the eye’s near point D′.

• Lens formula (Cartesian sign convention): 1/f = 1/v − 1/u, where u is object distance (negative for real objects), v is image distance (negative for virtual image on object side).

• Power of lens: P (in dioptre) = 1/f (in metres). For convex lens, f > 0, P > 0.

• For reading at 25 cm: take u = −25 cm; set v = −D′ (in cm). Then find f and P.

Solved Numerical :

Problem 1: A person’s near point is 50 cm. What power of convex lens is needed so that he can read at 25 cm?

Solution:

Given: D′ = 50 cm, u = −25 cm, v = −50 cm.

Lens formula: 1/f = 1/v − 1/u = 1/(−50) − 1/(−25) = −0.02 + 0.04 = 0.02 (in cm⁻¹).

Thus f = 50 cm = 0.50 m → P = 1/f = 1/0.50 = +2.0 D.

Answer: +2.0 dioptre convex lens.

 

Problem 2: A hypermetropic eye has near point at 1.0 m. Find the focal length and power of the correcting lens for comfortable reading at 25 cm.

Solution:

Given: D′ = 100 cm, u = −25 cm, v = −100 cm.

1/f = 1/v − 1/u = 1/(−100) − 1/(−25) = −0.01 + 0.04 = 0.03 (cm⁻¹).

f = 33.33… cm = 0.333… m → P ≈ 1/0.333… ≈ +3.0 D.

Answer: f ≈ +33.3 cm, P ≈ +3.0 D.

 

Problem 3: A person can read only if the book is at or beyond 80 cm from the eye. What power of spectacles is required to enable reading at 25 cm?

Solution:

Given: D′ = 80 cm, u = −25 cm, v = −80 cm.

1/f = 1/v − 1/u = 1/(−80) − 1/(−25) = −0.0125 + 0.04 = 0.0275 (cm⁻¹).

f = 36.36… cm = 0.3636… m → P ≈ 1/0.3636… ≈ +2.75 D.

Answer: ≈ +2.75 D.

 

Problem 4: What is the near point of a hypermetropic person who uses +2.5 D spectacles for reading at 25 cm? Assume thin lens and small vertex distance.

Solution:

Given: P = +2.5 D → f = 1/P = 0.4 m = 40 cm.

For reading at 25 cm, take u = −25 cm, f = +40 cm. Solve for v: 1/f = 1/v − 1/u.

1/40 = 1/v − (−1/25) ⇒ 1/v = 1/40 − 1/25 = (5 − 8)/200 = −3/200 ⇒ v = −200/3 ≈ −66.7 cm.

Thus D′ ≈ 66.7 cm (the virtual image is formed at the eye’s near point).

Answer: Near point ≈ 66.7 cm.

 

Problem 5: A person’s near point is at 60 cm. He wants to read at 30 cm (instead of the usual 25 cm). What power is required?

Solution:

Given: object distance u = −30 cm (chosen reading distance), v = −60 cm (near point).

1/f = 1/v − 1/u = 1/(−60) − 1/(−30) = −0.0167 + 0.0333 = 0.0167 (cm⁻¹).

f = 60 cm = 0.60 m → P = 1/0.60 ≈ +1.67 D.

Answer: ≈ +1.67 D.

 

Problem 6: A hypermetropic person uses spectacles of focal length +25 cm. At what distance from the eye can he see clearly without strain (i.e., what is his near point)? Assume reading at 25 cm.

Solution:

Given: f = +25 cm. For reading at 25 cm, u = −25 cm.

1/f = 1/v − 1/u ⇒ 1/25 = 1/v − (−1/25) ⇒ 1/v = 1/25 − 1/25 = 0 ⇒ v → −∞.

This means the person’s near point is effectively at infinity without accommodation; with this lens, an object at 25 cm appears at infinity.

Answer: The lens relaxes accommodation by forming the image at infinity; near point (effective for comfortable viewing) is ∞.

 

Problem 7: A student’s near point is 40 cm. (a) Find the power needed for reading at 25 cm. (b) If the same lens is used to view an object at 20 cm, where will the image form for the eye?

Solution:

(a) Use u = −25 cm, v = −40 cm ⇒ 1/f = 1/(−40) − 1/(−25) = −0.025 + 0.04 = 0.015 (cm⁻¹).

f = 66.7 cm = 0.667 m → P ≈ +1.50 D.

(b) For the same lens, f = +66.7 cm. With u = −20 cm: 1/f = 1/v − 1/u ⇒ 1/66.7 = 1/v − (−1/20).

Compute: 1/v = 1/66.7 − (−1/20) = 0.015 − (−0.05) = 0.065 (cm⁻¹) ⇒ v ≈ 15.38 cm (positive).

But positive v would indicate a real image on the far side, which a simple reading lens cannot deliver for such geometry; in fact, the thin-lens assumption with small f/u here suggests the eye cannot accommodate this; practical takeaway: object at 20 cm is too close—blurry without additional accommodation.

Answer: (a) +1.50 D. (b) The combination won’t give a comfortable virtual image at the eye’s near point; reading at 20 cm is not supported with this lens.

 

Problem 8: A person has near point at 1.5 m. What is the minimum power of the lens so that he can read a book at 25 cm?

Solution:

Given:  D′ = 150 cm, u = −25 cm, v = −150 cm.

1/f = 1/v − 1/u = 1/(−150) − 1/(−25) = −0.00667 + 0.04 = 0.03333 (cm⁻¹).

f = 30 cm = 0.30 m → P = 1/0.30 ≈ +3.33 D.

Answer: ≈ +3.33 D.

 

Problem 9: A hypermetropic person can barely see objects at 60 cm. If he puts on +2.0 D glasses, at what nearest distance can he read comfortably (assume image at near point for relaxed reading)?

Solution:

Given:  D′ = 60 cm; P = +2.0 D ⇒ f = 0.5 m = 50 cm.

We set v = −D′ = −60 cm and f = +50 cm, solve for u (reading distance).

1/f = 1/v − 1/u ⇒ 1/50 = 1/(−60) − 1/u ⇒ 1/u = 1/(−60) − 1/50.

Compute: 1/u = −0.0167 − 0.02 = −0.0367 (cm⁻¹) ⇒ u ≈ −27.2 cm.

Answer: ≈ 27 cm from the eye.

 

Problem 10: A person wants to shift his comfortable reading distance from 40 cm to the normal 25 cm using spectacles. Find the required power.

Solution:

Given: D′ = 40 cm; u = −25 cm; v = −40 cm.

1/f = 1/(−40) − 1/(−25) = −0.025 + 0.04 = 0.015 (cm⁻¹).

f = 66.7 cm = 0.667 m → P ≈ +1.50 D.

Answer: +1.50 D.

Practice Problems (Try Yourself)

1) Near point = 1.2 m. Power needed to read at 25 cm?

2) Near point = 75 cm. What focal length and power correct for 25 cm reading?

3) A person uses +3.0 D spectacles. If he reads at 25 cm, what is his near point?

4) With +2.25 D glasses, what is the closest reading distance for a person whose near point is 70 cm?

5) A student wants to read at 30 cm but has near point at 90 cm. Find the power required.

6) Near point unknown. Person can read at 25 cm using a +2.0 D lens. Find his near point.

7) A hypermetropic person wants the image to form at infinity while reading at 25 cm. What power is needed?

8) A person’s near point is 45 cm. If he accidentally uses a +1.0 D lens, what will be the closest distance at which he can read (image at near point)?

9) A +2.5 D lens is replaced by a +2.0 D lens. For reading at 25 cm, by how much does the near point (of the eye) implied by the lens change?

10) A person can read only at 1.8 m. What power of lens is required for 25 cm reading?

Answer Key (Practice)

1) +3.33 D (f ≈ +30 cm).

2) f ≈ +42.9 cm, P ≈ +2.33 D.

3) f = 1/P = 0.333 m = 33.3 cm → D′ ≈ 50 cm (using u = −25 cm).

4) ≈ 29.2 cm (solve with D′ = 70 cm, f from P = +2.25 D).

5) ≈ +2.22 D (u = −30 cm, v = −90 cm).

6) D′ = 50 cm (since f = 0.5 m, u = −25 cm).

7) P = +4.0 D (f = +0.25 m so that object at 25 cm is imaged at infinity).

8) ≈ 83.3 cm.

9) Change from ≈ 66.7 cm to 50 cm; near point decreases by ≈ 16.7 cm.

10) ≈ +3.78 D.

3) Presbyopia:

Definition:

Presbyopia is a common defect of vision in which an aged person cannot see nearby objects clearly. It usually appears after the age of 40–45 years and gradually increases with age.

Causes of Presbyopia

• With age, the ciliary muscles of the eye become weak.
• The flexibility of the eye lens decreases.
• As a result, the eye loses its power of accommodation (ability to adjust its focal length).
• The eye cannot focus clearly on nearby objects → blurred vision.

Symptoms:

1. Difficulty in reading small print (especially in dim light).
2. Holding a book or newspaper farther away to read.
3. Eyestrain or headache while doing close work (like stitching, reading, or mobile usage).

Correction of Presbyopia

• Presbyopia is corrected by using convex (converging) lenses in spectacles.
• If the person also has myopia (short-sightedness) along with presbyopia, then bifocal lenses are used:

                         - Upper part → concave lens (for distant vision).

                         - Lower part → convex lens (for near vision).

Difference between Presbyopia an Hypermetropia :

Feature	Presbyopia	Hypermetropia
Age group	Old age (40+ years)	Any age (even children)
Cause	Weak ciliary muscles & reduced flexibility of lens.	Short eye-ball or low converging power of lens
Correction	Bifocal Lenses	Convex Lenses

Key Point for Exams: 
        Presbyopia is not the same as hypermetropia; it is age-related and due to weakening of eye muscles and reduced lens flexibility.

Refraction of Light through a Prism:

What is a Prism?

• A prism is a transparent refracting optical element with two plane polished surfaces inclined at an angle (called the angle of prism).
• In Class 10, we generally study a triangular glass prism.

Refraction in a Prism:

When a ray of light passes through a prism, it undergoes two refractions:
1. First Refraction: When light enters from air into the glass prism (rarer → denser medium). The ray bends towards the normal.
2. Second Refraction: When light exits from glass into air (denser → rarer medium). The ray bends away from the normal.

Thus, the emergent ray is not in the same direction as the incident ray.

Angle of Deviation (D) :The angle between the incident ray (extended forward) and the emergent ray is called the Angle of Deviation (D).

Important Terms

1. Incident Ray (NP): The incoming ray striking the first refracting surface of the prism.
2. Refracted Ray (PQ): The ray inside the prism after the first refraction.
3. Emergent Ray (QM): The ray emerging out of the prism after the second refraction.
4. Angle of Prism (A): The angle between the two plane refracting surfaces of the prism.
5. Angle of Deviation (D): The angle between the direction of the incident ray and the emergent ray.

Relation between angle of prism (A), angle of deviation (D), angle of incidence (i₁) and angle of emergence ( i₂) :

i₁ +  i₂   = A + D

Where,

•  i₁= angle of incidence
•  i₂ = angle of emergence 
•  A = angle of prism 
•  D = angle of deviation

Dispersion of Light Through a Prism:

White light consists of different wavelengths (colors). When white light passes through a prism, each color bends by a different amount because refractive index depends on wavelength. Violet bends the most, Red bends the least. This spreading of white light into its component colors is called dispersion. The band of colors obtained is known as the spectrum.

Applications

1. Formation of Rainbow – natural dispersion by water droplets acting as prisms.
2. Spectrometers – to determine refractive index.
3. Prism in optics – used in binoculars, periscopes, cameras.

Cause of Dispersion:

Cause of Dispersion of Light:

Definition: Dispersion of light is the phenomenon of splitting of white light into its constituent seven colors (Violet, Indigo, Blue, Green, Yellow, Orange, Red – VIBGYOR) when it passes through a transparent medium such as a glass prism.

Example: When sunlight passes through a prism, it forms a beautiful spectrum.

Why does Dispersion Happen?

Dispersion occurs because different colors of light travel at different speeds inside a medium (like glass or water).

• In air or vacuum: All colors of light travel with the same speed (3 × 10⁸ m/s). Hence, no dispersion occurs in air/vacuum.

• In glass or water: The speed of light depends on its wavelength (color). Each color has a different refractive index (μ) inside the medium.

Refractive Index Dependence

Refractive index is given by: μ = c/v

where, c = speed of light in vacuum, v = speed of light in the medium.

Since v (speed) changes with wavelength, μ (refractive index) also changes with wavelength.

In glass:

• Violet light has shortest wavelength → slowed down the most → highest refractive index.

• Red light has longest wavelength → slowed down the least → lowest refractive index.

Thus, violet bends the most, and red bends the least inside the prism.

Summary of Cause of Dispersion

• Dispersion occurs due to variation of refractive index of the medium with wavelength of light.

• This variation is called chromatic dispersion.

• Since each color is refracted (bent) by a different amount, the white light spreads into a spectrum.

Order of Bending (VIBGYOR)

• Maximum bending → Violet (short wavelength, high μ).

• Minimum bending → Red (long wavelength, low μ).

• Order: V < I < B < G < Y < O < R in increasing wavelength, and opposite in bending.

Key Point for Exams

Cause of Dispersion: The dispersion of white light by a prism occurs because the refractive index of the prism material is different for different wavelengths (colors) of light. Hence, each color is deviated by a different angle, splitting the white light into a spectrum.

Rainbow Formation

Introduction

• A rainbow is a natural optical phenomenon caused by the dispersion, refraction, and reflection of sunlight in water droplets present in the atmosphere. 
• It usually appears in the sky opposite to the Sun, after rainfall or near waterfalls/fountains when water droplets are suspended in the air.

Conditions for Rainbow Formation

• Presence of tiny water droplets in the air (after rain or near mist).
• Sunlight shining from behind the observer.
• The sky must be relatively dark opposite to the Sun for the rainbow to be visible clearly.

Process of Rainbow Formation

• Rainbow formation involves three key processes inside a raindrop:

1. Refraction (Bending of light)

• When sunlight enters a spherical water droplet, it slows down and bends because of change of medium (air → water). 
• The white light splits into its seven constituent colours (VIBGYOR) due to dispersion. Violet bends the most, and red bends the least.

2. Internal Reflection

• The refracted light strikes the inner surface of the raindrop. 
• Since water is denser than air, part of the light is reflected back inside the drop. This is called internal reflection.

3. Refraction (Again)

• The internally reflected light comes out of the droplet.
•  During this, it undergoes refraction again (water → air). The colours spread further apart, enhancing the dispersion effect.

Types of Rainbow

1. Primary Rainbow:-

• Formed by one internal reflection inside the water droplet.
• Red colour appears on the outer edge and violet on the inner edge.

2. Secondary Rainbow:-

• Formed by two internal reflections inside the water droplet.
• It is fainter and appears outside the primary rainbow.
• Colours are reversed (red inside, violet outside).

Order of Colours

• The rainbow consists of seven colours in the order:
• Violet, Indigo, Blue, Green, Yellow, Orange, Red (VIBGYOR).

Important Points

• Rainbow is seen in the sky opposite to the Sun.
• The observer’s back should face the Sun.
• Each raindrop contributes only one colour towards the rainbow for a given observer; the complete rainbow is formed by millions of droplets together.

Applications / Conceptual Understanding

• Rainbow is an example of dispersion of light by natural phenomena. 
• The same principle is used in prisms to split light into seven colours.

Atmospheric Refraction

Definition

• Atmospheric refraction is the bending of light as it passes through the Earth’s atmosphere due to the variation in the density of air layers (Troposphere, Stratosphere, Mesosphere, Thermosphere and Exosphere).
• Since the atmosphere is not uniform (it has varying temperatures and densities), light rays get refracted continuously.

Cause

Air density and refractive index change with altitude:
- At higher altitudes → air is rarer (less dense).
- Near the Earth’s surface → air is denser.
Light travelling through these varying layers bends towards the normal when entering denser layers.

Phenomena Explained by Atmospheric Refraction

1. Apparent Position of Stars

• Stars appear slightly higher in the sky than their actual position. 
• This is because starlight bends gradually towards the Earth as it enters denser air layers.

2. Twinkling of Stars

• The apparent position of stars keeps changing due to continuous changes in air density and temperature. 
• This makes stars appear to twinkle. 
• Planets do not twinkle because they are closer and appear as large discs, averaging out the variations.

3. Advanced Sunrise and Delayed Sunset

• We see the Sun about 2 minutes earlier before actual sunrise and about 2 minutes later after sunset. 
• Reason: Sunlight bends around the horizon due to refraction, making the Sun visible even when it is geometrically below the horizon.

4. Looming / Mirage-like Effects

• Objects near the horizon (like ships or trees) sometimes appear raised or distorted due to atmospheric refraction.

Important Points

• Refraction occurs at every layer of the atmosphere, but the bending is gradual so it seems like a smooth curve.
• The shift in apparent position is about 0.5° (half a degree).
• Atmospheric refraction is responsible for many everyday optical phenomena in the sky.

Scattering of Light

Definition

Scattering of light is the phenomenon in which light is redirected in different directions when it strikes very small particles (like dust, gas molecules, water droplets) suspended in a medium.

Cause

Scattering happens because the size of particles in the medium is comparable to or smaller than the wavelength of light.
- Shorter wavelengths (blue, violet) scatter more.
- Longer wavelengths (red, orange) scatter less.

Examples & Applications

1. Why is the Sky Blue?

Sunlight contains all colours. Molecules of air scatter shorter wavelength light (blue) more than red. Hence, we see the sky as blue.

2. Why are Sunrises and Sunsets Red?

At sunrise and sunset, sunlight travels a longer distance through the atmosphere. Most of the blue light is scattered away, so only red and orange (longer wavelength, least scattered) reach our eyes.

3. Tyndall Effect

The scattering of light by colloidal particles is called the Tyndall Effect.
Examples:
- Beam of sunlight in a dusty room looks visible due to scattering by dust.
- Headlights of a vehicle in fog.

4. Other Applications

- Danger signal lights are red because red light scatters the least and can travel the farthest.
- The white appearance of clouds is due to scattering of all wavelengths by large water droplets.

Important Points

- The intensity of scattered light is inversely proportional to the square of the wavelength (∝ 1/λ²).

- Shorter the wavelength → greater the scattering.

- Scattering explains natural phenomena like blue sky, red sunsets, white clouds, and visibility of beams in mist/fog.

Download : Human Eye and Colorful World Notes PDF

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