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14 public questions tagged with this topic.

Optical Phenomena and Applications

What is the intensity of light after passing through two polaroids with pass-axes at \( 45^\circ \), if the initial unpo

**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. After the first polaroid, I = (I₀/2) . After the second at 45° , I = (I₀/2) cos² 45° = (I₀/2) × (1/2) = (I₀/4) . 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 (I₀/4),

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

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Wave Properties, Frequency and Energy Conservation

What is the intensity at a point in a double-slit experiment where the phase difference is \( 3\pi \), if the maximum in

**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. Intensity I = 4I₀ cos²(Φ/2) . For Φ = 3π , I = 4I₀ cos²((3π/2)) = 4I₀ × 0 = 0 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n

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

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Interference - Double-Slit and Coherence

In a double-slit experiment, if the path difference at a point is \( \lambda/2 \), what is the resulting intensity if th

**Coherence** requires constant phase difference, two independent sources not coherent because phase random, prevents interference pattern, laser coherent, visibility of fringes determined by coherence, intensity and phase difference, coherent sources produce stable interference, incoherent intensity adds I = I₁+I₂ no interference, interference pattern distinguishes from diffraction. Intensity I = 4I₀ cos²(Φ/2) , where Φ = (2π/λ) Δ . For Δ = λ/2 , Φ = (2π/λ) · (λ/2) = π , so I = 4I₀ cos²(π/2) = 4I₀ × 0 = 0 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v,

Ref: NCERT > Physics Book > Wave Optics > Interference - Double-Slit and Coherence

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Polarization and Malus Law

What is the intensity of light after passing through two polaroids with pass-axes at \( 30^\circ \), if the initial unpo

**Polaroid rotation** intensity varies sinusoidally with angle due to Malus law I = I₀ cos²θ, when polaroid rotated 90° from initial, intensity goes from max to zero, for unpolarized light rotating polaroid does not change intensity after first polaroid because average, but second polaroid intensity depends on relative angle, explains why intensity transmitted through two polaroids drops to zero when perpendicular. After the first polaroid, I = (I₀/2) . After the second at 30° , I = (I₀/2) cos² 30° = (I₀/2) × (3/4) = (3I₀/8) . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀

Ref: NCERT > Physics Book > Wave Optics > Polarization and Malus Law

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Wavefront and Huygens Principle

Why does the intensity of transmitted light through a polaroid decrease when rotated, even if the incident light is unpo

**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. Unpolarized light becomes polarized after the first polaroid, and the second polaroid’s pass-axis alignment determines the transmitted component, reducing intensity as the angle increases. 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 = 2

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

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Interference - Double-Slit and Coherence

In a double-slit experiment, if the intensity at the central maximum is \( 4I_0 \), what is the intensity where the path

**Double-slit vs single-slit** double-slit interference pattern has equally spaced bright fringes with envelope due to single-slit diffraction, single-slit central maximum width 2λ D/a, intensity of secondary maxima decreases with order, condition for coherence constant frequency and phase, path difference for bright n λ, dark (n+½)λ. Intensity I = 4I₀ cos²(Φ/2) , where Φ = (2π/λ) Δ . For Δ = λ/4 , Φ = (2π/λ) · (λ/4) = (π/2) , I = 4I₀ cos²(π/4) = 4I₀ × (1/2) = 2I₀ . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC =

Ref: NCERT > Physics Book > Wave Optics > Interference - Double-Slit and Coherence

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Interference - Double-Slit and Coherence

What is the intensity of two incoherent light sources each of intensity \( I_0 \) when combined at a point?

**Coherence** requires constant phase difference, two independent sources not coherent because phase random, prevents interference pattern, laser coherent, visibility of fringes determined by coherence, intensity and phase difference, coherent sources produce stable interference, incoherent intensity adds I = I₁+I₂ no interference, interference pattern distinguishes from diffraction. For incoherent sources, intensities add directly, so I = I₀ + I₀ = 2I₀ . 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 2I₀, illustr

Ref: NCERT > Physics Book > Wave Optics > Interference - Double-Slit and Coherence

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Superposition, Resultant Amplitude and Intensity

What is the intensity at a point in a double-slit experiment where the phase difference is \( \pi/2 \), if the maximum i

**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) . For Φ = (π/2) , I = 4I₀ cos²((π/4)) = 4I₀ ((√(2)/2))² = 4I₀ × (1/2) = 2I₀ . 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 2I₀, illustrating interference, diffraction and polarization principles.

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

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Polarization and Malus Law

What is the intensity of light after passing through two polaroids with their pass-axes perpendicular to each other, if

**Polaroid rotation** intensity varies sinusoidally with angle due to Malus law I = I₀ cos²θ, when polaroid rotated 90° from initial, intensity goes from max to zero, for unpolarized light rotating polaroid does not change intensity after first polaroid because average, but second polaroid intensity depends on relative angle, explains why intensity transmitted through two polaroids drops to zero when perpendicular. After the first polaroid, intensity is (I₀/2) . For the second polaroid at 90° , I = (I₀/2) cos² 90° = (I₀/2) × 0 = 0 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I =

Ref: NCERT > Physics Book > Wave Optics > Polarization and Malus Law

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Polarization and Malus Law

What is the intensity of light after passing through two polaroids with pass-axes at \( 60^\circ \), if the initial unpo

**Two polaroids** with pass-axes perpendicular intensity zero because second blocks component, with 45° intensity I₀/4 if initial after first I₀, if initial unpolarized I₀, after first I₀/2 then after second at 45° I₀/4. Three polaroids first and third crossed 90°, middle at 45° maximum transmission because middle rotates polarization, I after first I₀/2, after middle at 45° I₀/4, after third at 45° to middle I₀/4×cos²45°= I₀/8, non-zero, maximum when middle 45°. After the first polaroid, I = (I₀/2) . After the second at 60° , I = (I₀/2) cos² 60° = (I₀/2) × (1/4) = (I₀/8) . Using Δ = d sinθ, y

Ref: NCERT > Physics Book > Wave Optics > Polarization and Malus Law

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Polarization and Malus Law

Why does the intensity of light transmitted through three polaroids reach a maximum when the middle one is at 45° to the

**Polarization** transverse wave property, light after polaroid polarized along pass-axis, Malus law I = I₀ cos²θ, θ angle between pass-axes, initial unpolarized intensity I₀ after first polaroid I₁ = I₀/2, after second at 45° I₂ = I₁ cos²45°= I₀/2×0.5= I₀/4, after two perpendicular 90° I=0 because cos90°=0, for 60° I= I₀/2×cos²60°= I₀/2×0.25= I₀/8. At 45°, the middle polaroid allows maximum projection of the polarized light from the first to pass through the third, optimizing the cosine-squared term in Malus’ law. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n,

Ref: NCERT > Physics Book > Wave Optics > Polarization and Malus Law

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Polarization and Malus Law

What is the intensity of light transmitted through two polaroids with their pass-axes at \( 45^\circ \) to each other, i

**Polaroid rotation** intensity varies sinusoidally with angle due to Malus law I = I₀ cos²θ, when polaroid rotated 90° from initial, intensity goes from max to zero, for unpolarized light rotating polaroid does not change intensity after first polaroid because average, but second polaroid intensity depends on relative angle, explains why intensity transmitted through two polaroids drops to zero when perpendicular. Using Malus’ law, I = I₀ cos² θ . For θ = 45° , cos 45° = (√(2)/2) , so I = I₀ ((√(2)/2))² = I₀ × (1/2) = (I₀/2) . Using Δ = d sinθ, y = n λ D/d, a sinθ =

Ref: NCERT > Physics Book > Wave Optics > Polarization and Malus Law

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