Q1. Fill in the blanks and rewrite the sentences

a. Refractive index depends on the velocity of light.

Explanation:

The refractive index (μ) of a medium is defined as the ratio of the speed of light in vacuum (or air) to the speed of light in that medium:

μ = c/v

Where:

  • μ = refractive index of the medium
  • c = speed of light in vacuum/air (3 × 10⁸ m/s)
  • v = speed of light in the medium

Since the refractive index is directly related to how much light slows down in a medium, it depends on the velocity of light in that medium. Different materials have different refractive indices because light travels at different speeds through them.

Example:

  • Air: μ ≈ 1.0003 (light travels almost at c)
  • Water: μ ≈ 1.33 (light slows to about 225,000 km/s)
  • Glass: μ ≈ 1.5 (light slows to about 200,000 km/s)
  • Diamond: μ ≈ 2.42 (light slows to about 124,000 km/s)

b. The change in direction of light rays while passing from one medium to another is called refraction.

Explanation:

Refraction is the bending of light when it passes from one transparent medium to another with a different optical density.

[Diagram showing light ray bending at air-water interface]

Fig: Light ray changing direction (refracting) at boundary between two media

Key points about refraction:

  1. Occurs due to change in speed of light in different media
  2. The bending happens at the interface between two media
  3. Light bends toward the normal when entering a denser medium (slower speed)
  4. Light bends away from the normal when entering a rarer medium (faster speed)
  5. The amount of bending depends on the refractive indices of both media
  6. Governed by Snell's Law: μ₁ sin θ₁ = μ₂ sin θ₂

Examples in daily life:

  • Straw appearing bent in a glass of water
  • Pool looking shallower than it actually is
  • Twinkling of stars
  • Mirages in deserts

Q2. Prove the following statements

A. If the angle of incidence and the angle of emergence of a light ray incident on a glass slab are i and e respectively, prove that i = e.

[Diagram of rectangular glass slab with light ray entering and emerging]

Fig: Rectangular glass slab PQRS showing refraction at both surfaces

Proof:

  1. Consider a rectangular glass slab PQRS with parallel surfaces.
  2. An incident ray AO strikes surface PQ at point O with angle of incidence ∠i = ∠AON (where NN′ is normal at O).
  3. When light travels from air (rarer medium, μ₁ ≈ 1) to glass (denser medium, μ₂ ≈ 1.5), it bends toward the normal according to Snell's Law.
  4. The refracted ray OB inside glass makes angle of refraction ∠r = ∠BON′.
  5. At surface SR, ray OB strikes at point B. Now light travels from glass (denser) to air (rarer), bending away from normal N₁N₁′.
  6. The emerging ray BC makes angle of emergence ∠e = ∠CBN₁.

Mathematical Proof using Snell's Law:

At first surface (air to glass):

μ_air × sin i = μ_glass × sin r ...(1)

At second surface (glass to air):

μ_glass × sin r′ = μ_air × sin e ...(2)

Since surfaces are parallel, the angle of refraction at first surface equals the angle of incidence at second surface:

r = r′

From equations (1) and (2):

μ_air × sin i = μ_air × sin e
sin i = sin e
∴ i = e

Conclusion: The incident ray and emergent ray are parallel to each other, and the lateral displacement depends on thickness of slab and refractive index.

Practical observation: When looking through a glass window, objects appear slightly shifted but not distorted because of this parallel emergence.

B. A rainbow is formed due to the combined effect of refraction, dispersion, and total internal reflection of light.

Proof and Explanation:

A rainbow is one of nature's most beautiful optical phenomena, appearing as a circular arc of colors in the sky after rainfall when sunlight is present.

[Diagram showing path of light through water droplet forming rainbow]

Fig: Light path through water droplet showing refraction, dispersion, and internal reflection

Step-by-step formation of rainbow:

1. Refraction

Sunlight enters a spherical water droplet in the atmosphere. As light passes from air (rarer) to water (denser), it bends toward the normal at the air-water interface.

2. Dispersion

White sunlight consists of seven colors (VIBGYOR) with different wavelengths. Since refractive index depends on wavelength (dispersion), each color bends by a different amount:

  • Violet bends most (shortest wavelength, highest refractive index)
  • Red bends least (longest wavelength, lowest refractive index)

This separation of colors is called dispersion.

3. Total Internal Reflection

After dispersion, the colored rays reach the inner surface of the water droplet. For the primary rainbow, light undergoes one total internal reflection at the back of the droplet.

Condition for total internal reflection: Light must travel from denser (water) to rarer (air) medium at an angle greater than critical angle (≈ 48.6° for water).

4. Refraction Again

After internal reflection, the dispersed light rays exit the droplet, undergoing refraction again (water to air), bending away from normal. This further separates the colors.

Types of Rainbows:

Type Internal Reflections Color Order Angular Radius
Primary Rainbow One Red outside, Violet inside 42° from antisolar point
Secondary Rainbow Two Violet outside, Red inside (reversed) 51° from antisolar point

Why circular? Rainbows are circular because they form at specific angles (42° for primary) from the line opposite the Sun (antisolar point). We see an arc because the ground cuts off the lower part.

Conditions for rainbow formation:

  • Sun must be behind the observer
  • Raindrops must be in front of the observer
  • Sun should be low in sky (early morning or late afternoon best)
  • Each observer sees their own personal rainbow from different droplets

Q3. Mark the correct answer

A. What is the reason for the twinkling of stars?

A. Explosions occurring in stars repeatedly

B. Absorption of light in the Earth's atmosphere

C. Movement of stars

D. Continuous change in refractive index of atmospheric gases

Correct Answer: D. Continuous change in refractive index of atmospheric gases

Detailed Explanation:

The twinkling of stars (scientific term: stellar scintillation) is caused by atmospheric refraction and turbulence.

[Diagram showing starlight refracting through turbulent atmospheric layers]

Fig: Starlight passing through turbulent atmosphere causing apparent position changes

How it works:

  1. Starlight enters Earth's atmosphere from space (vacuum).
  2. The atmosphere has layers of air with varying temperature, density, and pressure.
  3. These variations cause continuous changes in the refractive index of air.
  4. As light passes through these turbulent layers, it undergoes continuous refraction in random directions.
  5. This causes:
    • Apparent change in star's position (wandering)
    • Change in brightness (twinkling)
    • Color separation sometimes (like a prism effect)
  6. The effect is more pronounced near the horizon where light travels through more atmosphere.

Why planets don't twinkle (much):

  • Planets are closer and appear as extended sources (discs), not point sources
  • Refraction effects from different points on the disc average out
  • Stars are so far away they appear as perfect point sources

Applications: This atmospheric turbulence is why astronomers prefer space telescopes (like Hubble) and use adaptive optics in ground telescopes.

B. We can see the Sun even when it is slightly below the horizon due to

A. Reflection of light

B. Refraction of light

C. Dispersion of light

D. Absorption of light

Correct Answer: B. Refraction of light

Detailed Explanation:

This phenomenon is called atmospheric refraction and causes the Sun (and Moon) to be visible even when they are geometrically below the horizon.

[Diagram showing atmospheric refraction making Sun appear higher]

Fig: Light from Sun below horizon bends in atmosphere, making it visible

How it works:

  1. The Earth's atmosphere is denser near the surface and gradually becomes rarer with height.
  2. When sunlight enters the atmosphere at a shallow angle (near sunrise/sunset), it passes through many layers of varying density.
  3. Light bends continuously toward the denser layers (toward the normal at each interface).
  4. This continuous refraction causes the light path to curve, following the Earth's curvature.
  5. As a result, we see the Sun in a position higher than its actual geometric position.
  6. The apparent shift is about 0.5° (approximately the Sun's diameter), allowing us to see it for extra 2 minutes at sunrise and sunset.

Effects:

  • Advanced sunrise: Sun appears 2 minutes before actual sunrise
  • Delayed sunset: Sun remains visible 2 minutes after actual sunset
  • Sun appears flattened at horizon due to different refraction of top and bottom edges
  • Same effect causes stars to appear higher than their actual positions

Mathematical aspect: The refractive index of air near surface is about 1.0003, decreasing to 1 at top of atmosphere. This gradient causes continuous bending.

C. If the refractive index of glass with respect to air is 3/2, what is the refractive index of air with respect to glass?

Correct Answer: Refractive index of air with respect to glass = 2/3

Solution with Explanation:

Given: Refractive index of glass with respect to air = μ_glass/air = 3/2

The refractive index of medium 2 with respect to medium 1 is defined as:

μ₂₁ = μ₂/μ₁ = v₁/v₂

Where μ₁ and μ₂ are absolute refractive indices, v₁ and v₂ are speeds of light in medium 1 and 2 respectively.

Step 1: Understand the given

μ_glass/air = μ_glass/μ_air = 3/2

Step 2: Find reciprocal

The refractive index of air with respect to glass is the reciprocal:

μ_air/glass = μ_air/μ_glass = 1 / (μ_glass/μ_air) = 1 / (3/2) = 2/3

Step 3: Physical interpretation

This makes physical sense because:

  • When light goes from glass to air, it speeds up (air is optically rarer)
  • The refractive index for rarer to denser is greater than 1
  • The refractive index for denser to rarer is less than 1
  • 2/3 ≈ 0.667 < 1, consistent with glass being denser than air

General formula: If μ_ba = x, then μ_ab = 1/x

Verification using speed of light:

If μ_glass/air = 3/2, then:

v_air / v_glass = 3/2 ⇒ v_glass = (2/3) v_air

Then μ_air/glass = v_glass / v_air = (2/3 v_air) / v_air = 2/3 ✓

Q4. Solve the following examples

A. If the speed of light in a medium is 1.5 × 10⁸ m/s, find the absolute refractive index of the medium.

Solution:

Given:

  • Speed of light in the medium: v = 1.5 × 10⁸ m/s
  • Speed of light in vacuum/air: c = 3 × 10⁸ m/s (standard value)

Formula: Absolute refractive index (μ) = Speed of light in vacuum / Speed of light in medium

μ = c / v

Calculation:

μ = (3 × 10⁸ m/s) / (1.5 × 10⁸ m/s)
μ = 3 / 1.5
μ = 2

Answer: The absolute refractive index of the medium is 2.

Physical interpretation:

  • Light travels twice as fast in vacuum/air as in this medium
  • The medium is optically denser than many common materials (water μ=1.33, glass μ=1.5)
  • Possible materials with μ≈2: Diamond (2.42), some types of glass, certain crystals
  • When light enters this medium from air, it will bend significantly toward the normal

Check using Snell's Law: If light enters this medium from air at 30° incidence:

1 × sin 30° = 2 × sin r
0.5 = 2 × sin r ⇒ sin r = 0.25 ⇒ r ≈ 14.5°

Light bends from 30° to 14.5° (toward normal), as expected for denser medium.

B. If the absolute refractive indices of glass and water are 3/2 and 4/3 respectively, find the refractive index of glass with respect to water.

Solution:

Given:

  • Absolute refractive index of glass: μ_g = 3/2 = 1.5
  • Absolute refractive index of water: μ_w = 4/3 ≈ 1.333

To find: Refractive index of glass with respect to water (μ_gw)

Formula: Refractive index of medium 2 with respect to medium 1:

μ₂₁ = μ₂ / μ₁

Here, medium 2 = glass, medium 1 = water

μ_gw = μ_g / μ_w

Calculation:

μ_gw = (3/2) ÷ (4/3)
μ_gw = (3/2) × (3/4)
μ_gw = 9/8

Decimal value: 9/8 = 1.125

Answer: The refractive index of glass with respect to water is 9/8 or 1.125.

Physical interpretation:

  • Since μ_gw = 1.125 > 1, glass is optically denser than water
  • When light passes from water to glass, it bends toward the normal
  • When light passes from glass to water, it bends away from the normal
  • The bending is less dramatic than from air to glass because the refractive index difference is smaller

Verification using speeds:

If μ_g = c/v_g = 1.5 and μ_w = c/v_w = 1.333, then:

v_g = c/1.5, v_w = c/1.333
μ_gw = v_w/v_g = (c/1.333) / (c/1.5) = 1.5/1.333 = 1.125 ✓

Application: This calculation is important in designing underwater optical instruments and understanding how light behaves in multi-medium systems.

Additional Practice Problem: If light travels from water to glass with the above refractive indices, and the angle of incidence in water is 40°, what is the angle of refraction in glass?

Solution: Using Snell's Law: μ_w sin i = μ_g sin r

Since we have μ_gw = μ_g/μ_w = 9/8, we can also use: sin i / sin r = μ_gw

sin 40° / sin r = 9/8
0.6428 / sin r = 1.125
sin r = 0.6428 / 1.125 ≈ 0.5714
r ≈ sin⁻¹(0.5714) ≈ 34.8°

Light bends from 40° to 34.8° (toward normal) when going from water to glass.

Key Concepts in Refraction of Light

Snell's Law

μ₁ sin θ₁ = μ₂ sin θ₂

Where:

  • μ₁, μ₂ = refractive indices of media 1 and 2
  • θ₁ = angle of incidence (in medium 1)
  • θ₂ = angle of refraction (in medium 2)

Remember: Angles are measured from the normal.

Critical Angle

sin C = μ_rarer / μ_denser

For total internal reflection:

  1. Light must travel from denser to rarer medium
  2. Angle of incidence > critical angle

Example: Water to air: C = sin⁻¹(1/1.33) ≈ 48.6°

Optical Phenomenon Caused by Example
Twinkling of stars Atmospheric refraction Stars appear to twinkle at night
Advanced sunrise/Delayed sunset Atmospheric refraction Sun visible before actual sunrise
Rainbow Refraction, dispersion, TIR After rain with sunlight
Mirage Total internal reflection In deserts on hot days
Sparkling of diamond Total internal reflection Diamond appears bright with facets
Bent pencil in water Refraction Pencil in glass of water