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#wave optics

40 public questions tagged with this topic.

Optical Phenomena and Applications

What is the condition for the third secondary maximum in a single-slit diffraction pattern?

**Convex lens focusing** plane wave into point because lens introduces phase delay proportional to thickness, converting plane wavefront to spherical converging to focal point, property ensures rays parallel to axis meet at focus, spherical aberration minimized for paraxial rays, lensmaker's formula determines focal length. Secondary maxima occur at θ ≈ ((n + (1/2))λ/a) . For the third secondary maximum, n = 3 , θ ≈ (7λ/2a) . 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 θ = (7λ/2a),

Ref: NCERT > Physics Book > Wave Optics > Optical Phenomena and Applications

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Wave Properties, Frequency and Energy Conservation

Why does the interference pattern from two slits disappear if the slits are too far apart?

**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. Large slit separation reduces the overlap of wavefronts, disrupting the consistent path difference needed for stable interference. Using Δ

Ref: NCERT > Physics Book > Wave Optics > Wave Properties, Frequency and Energy Conservation

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Wave Properties, Frequency and Energy Conservation

What causes the intensity of light to be zero at certain points in a diffraction pattern?

**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. Complete destructive interference occurs when secondary wavelets from different parts of the slit cancel each other out, resulting in zero intensity at minima. 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 Destructive interference, illustrating

Ref: NCERT > Physics Book > Wave Optics > Wave Properties, Frequency and Energy Conservation

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Wave Properties, Frequency and Energy Conservation

What is the condition for the fifth minimum in a single-slit diffraction pattern?

**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. Minima occur at sin θ = (nλ/a) . For the fifth minimum, n = 5 , so θ = sin⁻¹((5λ/a)) . 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 θ = (5λ/a), illustrating interference, diffraction and polarization principles.

Ref: NCERT > Physics Book > Wave Optics > Wave Properties, Frequency and Energy Conservation

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Superposition, Resultant Amplitude and Intensity

What is the intensity of light in the photon picture, according to the wave optics concept?

**Intensity at point** in double-slit I = I_max cos²(φ/2), φ = (2π/λ)Δ, for Δ=λ/4 φ=π/2 I= I_max cos²(π/4)= I_max/2 =2I₀, for Δ=λ/3 φ=2π/3 I= I_max cos²(π/3)= I_max×0.25= I₀, for Δ=5λ/8 φ=5π/4? Actually φ=2π×5/8=5π/4, cos²(5π/8)=?, path difference for destructive φ=(2n+1)π, constructive 2nπ. In the photon picture, intensity is determined by the number of photons crossing a unit area per unit time. 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 Number of photons per unit area per unit time, illustrating interferenc

Ref: NCERT > Physics Book > Wave Optics > Superposition, Resultant Amplitude and Intensity

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Refraction, Refractive Index and Critical Angle

What determines the intensity of light in the wave picture?

**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. In the wave picture, intensity is proportional to the square of the amplitude of the wave. 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 Square of the amplitude, illustrating interference, dif

Ref: NCERT > Physics Book > Wave Optics > Refraction, Refractive Index and Critical Angle

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Wavefront and Huygens Principle

Why does the interference pattern from two slits require light to originate from a single source?

**Wavefront types** point source spherical, distant point source plane, after convex lens plane wave focuses to point because lens adds phase delay proportional to thickness, converging spherical wavefront, after concave mirror plane wave becomes spherical converging to focus, illustrating Huygens construction. A single source ensures coherence, maintaining a constant phase relationship between the waves from the slits, essential for a stable interference pattern. 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 To

Ref: NCERT > Physics Book > Wave Optics > Wavefront and Huygens Principle

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Superposition, Resultant Amplitude and Intensity

What is the intensity at a point in a double-slit experiment where the path difference is \( 5\lambda/2 \), if the maxim

**Intensity at point** in double-slit I = I_max cos²(φ/2), φ = (2π/λ)Δ, for Δ=λ/4 φ=π/2 I= I_max cos²(π/4)= I_max/2 =2I₀, for Δ=λ/3 φ=2π/3 I= I_max cos²(π/3)= I_max×0.25= I₀, for Δ=5λ/8 φ=5π/4? Actually φ=2π×5/8=5π/4, cos²(5π/8)=?, path difference for destructive φ=(2n+1)π, constructive 2nπ. Intensity I = 4I₀ cos²(Φ/2) , where Φ = (2π/λ) Δ . For Δ = (5λ/2) , Φ = (2π/λ) · (5λ/2) = 5π , I = 4I₀ cos²((5π/2)) = 4I₀ × 0 = 0 . 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

Ref: NCERT > Physics Book > Wave Optics > Superposition, Resultant Amplitude and Intensity

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Superposition, Resultant Amplitude and Intensity

What is the intensity at a point in a double-slit experiment where the path difference is \( 7\lambda/2 \), if the maxim

**Superposition principle** resultant displacement sum of individual, for two coherent waves amplitude a each, resultant amplitude A = √(a² + a² +2a² cosφ)=2a|cos(φ/2)|, phase difference φ, path difference Δ = (φ/2π)λ, for φ=π/2 A=√2 a, for φ=6π cos3π=-1? Actually φ=6π cos3π? A=2a|cos3π|=2a, for φ=4π A=2a, intensity I ∝ A², maximum I_max=4I₀ when φ=0, I=2I₀(1+cosφ)=4I₀ cos²(φ/2). Intensity I = 4I₀ cos²(Φ/2) , where Φ = (2π/λ) Δ . For Δ = (7λ/2) , Φ = (2π/λ) · (7λ/2) = 7π , I = 4I₀ cos²((7π/2)) = 4I₀ × 0 = 0 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n

Ref: NCERT > Physics Book > Wave Optics > Superposition, Resultant Amplitude and Intensity

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Wavefront and Huygens Principle

What is the angular position of the third minimum in a single-slit diffraction pattern if the slit width is \( 4.0 \, \m

**Wavefront types** point source spherical, distant point source plane, after convex lens plane wave focuses to point because lens adds phase delay proportional to thickness, converging spherical wavefront, after concave mirror plane wave becomes spherical converging to focus, illustrating Huygens construction. Minima occur at sin θ = (nλ/a) . For the third minimum, n = 3 . λ = 4.0 × 10⁻⁷ m , a = 4.0 × 10⁻⁶ m . sin θ = (3 × 4.0 × 10⁻⁷/4.0 × 10⁻⁶) = 0.3 , θ = sin⁻¹(0.3) ≈ 17.5° . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ,

Ref: NCERT > Physics Book > Wave Optics > Wavefront and Huygens Principle

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Wavefront and Huygens Principle

Why does the interference pattern from two slits vanish if one slit is covered?

**Wavefront types** point source spherical, distant point source plane, after convex lens plane wave focuses to point because lens adds phase delay proportional to thickness, converging spherical wavefront, after concave mirror plane wave becomes spherical converging to focus, illustrating Huygens construction. Interference requires superposition from two sources; covering one slit eliminates the second wave, leaving only a diffraction pattern from the single slit. 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 On

Ref: NCERT > Physics Book > Wave Optics > Wavefront and Huygens Principle

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Wavefront and Huygens Principle

What is the condition for destructive interference in a double-slit experiment?

**Wavefront types** point source spherical, distant point source plane, after convex lens plane wave focuses to point because lens adds phase delay proportional to thickness, converging spherical wavefront, after concave mirror plane wave becomes spherical converging to focus, illustrating Huygens construction. Destructive interference occurs when the path difference is an odd multiple of half the wavelength, i.e., Δ = (n + (1/2))λ . 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 Path difference = (n + (1/2))λ, il

Ref: NCERT > Physics Book > Wave Optics > Wavefront and Huygens Principle

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