Wave Nature of Light: Hygens' Principle
ICSE · Class 12 · Physics
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Monochromatic light of wavelength 600 nm travels from air into a glass medium of refractive index 1.5. What is the wavelength of light inside glass and what happens to its frequency?
Using Huygens' principle, the proof of Snell's law gives sin i / sin r = v₁/v₂. If the angle of incidence is 45° and the angle of refraction is 30°, what is the refractive index of the second medium with respect to the first?
A ray of light travels a distance of 3 cm in glass (n = 1.5) and then 4 cm in water (n = 1.33). What is the total optical path length?
For the same angle of incidence (60°), angles of refraction in media P, Q, and R are 30°, 45°, and 20° respectively. Which medium has the HIGHEST refractive index?
Sample Questions
According to Huygens' wave theory, how does the refractive index relate to the speed of light in two media? If light travels at 2.0 × 10⁸ m/s in medium 1 and 1.5 × 10⁸ m/s in medium 2, what is ₁n₂?
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₁n₂ = v₁/v₂ = 4/3 ≈ 1.33
Step 1: Huygens' principle proves that ₁n₂ = sin i / sin r = v₁/v₂ (speed in first medium / speed in second medium). Step 2: ₁n₂ = v₁/v₂ = (2.0 × 10⁸) / (1.5 × 10⁸) = 2.0/1.5 = 4/3 ≈ 1.33. Step 3: This value (≈1.33) is the refractive index of water with respect to a medium like dense glass — physically meaningful. Step 4: Wrong option B inverts the ratio (that would be ₂n₁). Wrong option D uses c instead of v₁, which would give the absolute refractive index of medium 1 with respect to vacuum, not between the two media.
In Huygens' construction for refraction, AB is a plane wavefront in medium 1 (speed v₁). After time t, edge B reaches point A' on the refracting surface. If the secondary wavelet from A in medium 2 has radius AB' = v₂t and BA' = v₁t, which triangle relationship correctly gives sin i?
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sin i = BA'/AA' from triangle BAA'
Step 1: In the Huygens' refraction diagram, AA' is the refracting surface. Triangle BAA' has BA' as the perpendicular from B to AA' (the incident wavefront is AB, so BA' ⊥ AA'). Step 2: The angle of incidence i is the angle between the incident ray (normal to AB) and the normal to SS'. In triangle BAA', angle BAA' = i, and sin(BAA') = BA'/AA'. Step 3: Similarly, in triangle AB'A', angle AA'B' = r, so sin r = AB'/AA'. Step 4: Dividing: sin i/sin r = BA'/AB' = v₁t/v₂t = v₁/v₂, giving Snell's law. Wrong option B gives sin r, not sin i. Wrong option C inverts the sine to give cosine-like ratio.
Light of frequency 6 × 10¹⁴ Hz travels from vacuum into a medium of refractive index 1.6. What is the speed of light in the medium? (c = 3 × 10⁸ m/s)
Show answer
1.875 × 10⁸ m/s
Step 1: The speed of light in a medium is v = c/n, where n is the refractive index. Step 2: v = (3 × 10⁸ m/s) / 1.6 = 1.875 × 10⁸ m/s. Step 3: Note that the frequency (6 × 10¹⁴ Hz) is irrelevant here — it is given as a distractor to test understanding that speed depends only on the medium (refractive index), not on frequency (in this context). Step 4: Wrong option B (4.8 × 10⁸) comes from multiplying c × n instead of dividing. Wrong option C (3.0 × 10⁸) ignores the medium. Wrong option D confuses n with the answer directly.
A plane wavefront is incident on a convex lens. What type of wavefront emerges from the lens, and why does the centre of the wavefront travel faster than the edges through the lens?
Show answer
Spherical converging wavefront; centre travels through thicker glass (slower), so edges emerge first making the wavefront converge
Step 1: A convex lens is thicker at the centre and thinner at the edges. Step 2: Speed of light in glass (v = c/n) is less than in air (n > 1). So light slows down more in thicker regions. Step 3: The central portion of the wavefront spends more time in glass (travels slower there), while the edges spend less time in glass. Step 4: As a result, the edges of the wavefront emerge from the lens before the centre, causing the wavefront to curve inward — forming a spherical converging wavefront. Wrong option B incorrectly states edges are thicker — in a convex lens the centre is thicker. Wrong opti
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