#speed of light

Optical Phenomena and Applications

What is the refractive index of a medium if the speed of light in it is \( 1.8 \times 10^8 \, \text{m/s} \) and in vacuu

**Incoherent sources** intensity adds I = I₁+I₂, no interference pattern because phase random, two independent sources cannot produce stable interference because phase difference fluctuates rapidly, coherent sources required with constant phase, laser coherent, visibility of fringes requires coherence, degree of coherence determines contrast. Refractive index n = (c/v) . c = 3.0 × 10⁸ m/s , v = 1.8 × 10⁸ m/s . n = (3.0 × 10⁸/1.8 × 10⁸) ≈ 1.67 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a

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Optical Phenomena and Applications

What is the speed of light in a medium if its refractive index is 1.2 and the speed in vacuum is \( 3.0 \times 10^8 \, \

**Polarization requires transverse waves** because only transverse can have orientation perpendicular to propagation, longitudinal cannot be polarized, wave theory requires light transverse to explain polarization, polaroids transmit only component along pass-axis, unpolarized has random transverse orientations, after polaroid polarized. Speed in a medium v = (c/n) . Given n = 1.2 , c = 3.0 × 10⁸ m/s , v = (3.0 × 10⁸/1.2) = 2.5 × 10⁸ m/s . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation gives

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Wave Properties, Frequency and Energy Conservation

What is the speed of light in a medium with refractive index 1.25, given the speed in vacuum is \( 3.0 \times 10^8 \, \t

**Frequency of light** remains unchanged when refracts from air into water because frequency determined by source, energy E= h f conserved, speed decreases v= c/n, wavelength decreases λ'=v/f= λ/n, energy of wave proportional to amplitude² not speed, intensity I =½ c ε₀ E₀², energy not depend on speed directly, when speed decreases amplitude may change but energy conserved, interference redistributes energy, total energy same, bright regions gain from dark. Speed in a medium v = (c/n) . Given n = 1.25 , c = 3.0 × 10⁸ m/s , v = (3.0 × 10⁸/1.25) = 2.4 × 10⁸ m/s . Using Δ = d sinθ,

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Wave Properties, Frequency and Energy Conservation

What is the frequency of light with a wavelength of \( 680 \, \text{nm} \) in air, given the speed of light in air is \(

**Wave model predicts** light bends away from normal when entering rarer medium because speed increases, Snell's law n₁ sinθ₁ = n₂ sinθ₂, n₁>n₂ so sinθ₂>sinθ₁ θ₂>θ₁ away from normal, towards normal when denser, wavefront slows in denser, Huygens construction shows bending. Frequency nu = (c/λ) . λ = 6.8 × 10⁻⁷ m , c = 3.0 × 10⁸ m/s . nu = (3.0 × 10⁸/6.8 × 10⁻⁷) ≈ 4.41 × 10¹⁴ Hz . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a

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Refraction, Refractive Index and Critical Angle

What is the speed of light in a medium with refractive index 1.8, given the speed in vacuum is \( 3.0 \times 10^8 \, \te

**Frequency of light** does not change on refraction, c = f λ, when enters denser medium speed decreases, wavelength decreases λ' = v/f = c/(n f) = λ/n, frequency same, for λ=480 nm f=c/λ=3×10⁸/480×10⁻⁹=6.25×10¹⁴ Hz, for 630 nm f=4.76×10¹⁴ Hz, for 540 nm f=5.56×10¹⁴ Hz, refractive index determines bending. Speed in a medium v = (c/n) . Given n = 1.8 , c = 3.0 × 10⁸ m/s , v = (3.0 × 10⁸/1.8) ≈ 1.67 × 10⁸ m/s . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n,

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Refraction, Refractive Index and Critical Angle

What is the refractive index of a medium if the speed of light in it is \( 2.5 \times 10^8 \, \text{m/s} \) and in vacuu

**Frequency of light** does not change on refraction, c = f λ, when enters denser medium speed decreases, wavelength decreases λ' = v/f = c/(n f) = λ/n, frequency same, for λ=480 nm f=c/λ=3×10⁸/480×10⁻⁹=6.25×10¹⁴ Hz, for 630 nm f=4.76×10¹⁴ Hz, for 540 nm f=5.56×10¹⁴ Hz, refractive index determines bending. Refractive index n = (c/v) . c = 3.0 × 10⁸ m/s , v = 2.5 × 10⁸ m/s . n = (3.0 × 10⁸/2.5 × 10⁸) = 1.2 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ'

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Refraction, Refractive Index and Critical Angle

Why does the wave theory of light predict that the speed of light decreases when it bends towards the normal during refr

**Critical angle** sinC = n₂/n₁, n₁ denser, n₂ rarer 1 for air, for C=36.9° n₁=1/sin36.9°=1/0.6=1.67, for C=39° n=1/sin39°=1/0.629=1.59, for n=1.85 C=arcsin(1/1.85)=arcsin0.5405=32.7°, total internal reflection occurs only when light travels from denser to rarer and incidence > C, because Snell's law would require sinθ₂>1 impossible. The wave theory suggests that light slows down in a denser medium, causing the wavefront to bend towards the normal as it takes less time to travel a shorter path in the slower medium. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n,

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Refraction, Refractive Index and Critical Angle

What is the speed of light in glass if its refractive index is 1.6 and the speed of light in vacuum is \( 3.0 \times 10^

**Refractive index** n = c/v, c=3×10⁸ m/s vacuum, v speed in medium, for v=2.25×10⁸ m/s n=3/2.25=1.33, for v=1.8×10⁸ n=1.67, for n=1.3 v=3×10⁸/1.3=2.31×10⁸ m/s, for n=1.6 v=1.875×10⁸ m/s, for n=1.2 λ_medium = λ_vacuum/n =600/1.2=500 nm, for n=1.75 λ=700/1.75=400 nm, for n=1.5 λ=750/1.5=500 nm, wavelength in medium λ' = λ/n. Speed in a medium v = (c/n) . Given n = 1.6 , c = 3.0 × 10⁸ m/s , v = (3.0 × 10⁸/1.6) = 1.875 × 10⁸ m/s . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ'

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Refraction, Refractive Index and Critical Angle

What is the speed of light in a medium with a refractive index of 1.5, given the speed of light in vacuum is \( 3.0 \tim

**Frequency of light** does not change on refraction, c = f λ, when enters denser medium speed decreases, wavelength decreases λ' = v/f = c/(n f) = λ/n, frequency same, for λ=480 nm f=c/λ=3×10⁸/480×10⁻⁹=6.25×10¹⁴ Hz, for 630 nm f=4.76×10¹⁴ Hz, for 540 nm f=5.56×10¹⁴ Hz, refractive index determines bending. Refractive index n = (c/v) , where c is the speed in vacuum and v is the speed in the medium. Given n = 1.5 , v = (c/n) = (3.0 × 10⁸/1.5) = 2.0 × 10⁸ m/s . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I =

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Refraction, Refractive Index and Critical Angle

What is the speed of light in a medium with refractive index 1.4, given the speed in vacuum is \( 3.0 \times 10^8 \, \te

**Refractive index** n = c/v, c=3×10⁸ m/s vacuum, v speed in medium, for v=2.25×10⁸ m/s n=3/2.25=1.33, for v=1.8×10⁸ n=1.67, for n=1.3 v=3×10⁸/1.3=2.31×10⁸ m/s, for n=1.6 v=1.875×10⁸ m/s, for n=1.2 λ_medium = λ_vacuum/n =600/1.2=500 nm, for n=1.75 λ=700/1.75=400 nm, for n=1.5 λ=750/1.5=500 nm, wavelength in medium λ' = λ/n. Speed in a medium v = (c/n) . Given n = 1.4 , c = 3.0 × 10⁸ m/s , v = (3.0 × 10⁸/1.4) ≈ 2.14 × 10⁸ m/s . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ'

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Refraction, Refractive Index and Critical Angle

What is the refractive index of a medium if the speed of light in it is \( 2.4 \times 10^8 \, \text{m/s} \) and in vacuu

**Critical angle** sinC = n₂/n₁, n₁ denser, n₂ rarer 1 for air, for C=36.9° n₁=1/sin36.9°=1/0.6=1.67, for C=39° n=1/sin39°=1/0.629=1.59, for n=1.85 C=arcsin(1/1.85)=arcsin0.5405=32.7°, total internal reflection occurs only when light travels from denser to rarer and incidence > C, because Snell's law would require sinθ₂>1 impossible. Refractive index n = (c/v) . c = 3.0 × 10⁸ m/s , v = 2.4 × 10⁸ m/s , n = (3.0 × 10⁸/2.4 × 10⁸) = 1.25 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A =

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Refraction, Refractive Index and Critical Angle

What is the speed of light in a medium with refractive index 1.35, given the speed in vacuum is \( 3.0 \times 10^8 \, \t

**Refractive index** n = c/v, c=3×10⁸ m/s vacuum, v speed in medium, for v=2.25×10⁸ m/s n=3/2.25=1.33, for v=1.8×10⁸ n=1.67, for n=1.3 v=3×10⁸/1.3=2.31×10⁸ m/s, for n=1.6 v=1.875×10⁸ m/s, for n=1.2 λ_medium = λ_vacuum/n =600/1.2=500 nm, for n=1.75 λ=700/1.75=400 nm, for n=1.5 λ=750/1.5=500 nm, wavelength in medium λ' = λ/n. Speed in a medium v = (c/n) . Given n = 1.35 , c = 3.0 × 10⁸ m/s , v = (3.0 × 10⁸/1.35) ≈ 2.22 × 10⁸ m/s . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ'

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