Wave Optics Class 12 Physics NCERT Solutions

Step-by-step solution

1. 1Reflected light stays in air, so its speed and wavelength are unchanged: v = c = 3.0 × 10⁸ m s⁻¹, λ = 589 nm.
2. 2Frequency for both beams: ν = c/λ = (3.0 × 10⁸)/(589 × 10⁻⁹) = 5.093 × 10¹⁴ Hz.
3. 3For the refracted light: v′ = c/μ = (3.0 × 10⁸)/1.33 = 2.26 × 10⁸ m s⁻¹.
4. 4Refracted wavelength: λ′ = λ/μ = 589/1.33 = 443 nm (frequency ν stays 5.09 × 10¹⁴ Hz).
5. 5Answer to (a): reflected light — λ = 589 nm, ν = 5.09 × 10¹⁴ Hz, v = 3.0 × 10⁸ m s⁻¹.
6. 6Answer to (b): refracted light — λ′ = 443 nm, ν = 5.09 × 10¹⁴ Hz, v′ = 2.26 × 10⁸ m s⁻¹.

Step-by-step solution

1. 1(a) A point source radiates in all directions, so each wavefront is a sphere centred on the source.
2. 2(b) Rays leaving a source placed at the focus of a convex lens emerge parallel after refraction, so the wavefront is a plane (normal to the rays).
3. 3(c) Light from a distant star arrives as a spherical wave whose radius is effectively infinite at the Earth, so the intercepted portion is a plane wavefront.

Step-by-step solution

1. 1(a) v = c/μ = (3.0 × 10⁸)/1.5 = 2.0 × 10⁸ m s⁻¹.
2. 2(b) No — the refractive index depends on wavelength (dispersion). For a given material, μ is larger for shorter wavelengths.
3. 3Violet has the shorter wavelength, hence the larger index and the smaller speed — violet travels slower in the glass prism.

Step-by-step solution

1. 1The n-th bright fringe lies at yₙ = nλD/d. For n = 4: y₄ = 4λD/d = 1.2 cm = 1.2 × 10⁻² m.
2. 2λ = (y₄ · d)/(4D) = (1.2 × 10⁻² × 0.28 × 10⁻³)/(4 × 1.4).
3. 3λ = (3.36 × 10⁻⁶)/5.6 = 6.0 × 10⁻⁷ m = 600 nm.

Step-by-step solution

1. 1The resultant intensity is I = 4I₀ cos²(φ/2), with phase φ = (2π/λ) × (path difference).
2. 2At path difference λ: φ = 2π, cos²(π) = 1, so I = 4I₀ = K, i.e. I₀ = K/4.
3. 3At path difference λ/3: φ = 2π/3, so φ/2 = π/3 and I = 4I₀ cos²(π/3) = 4I₀ × (1/2)² = I₀.
4. 4I = K/4.

Step-by-step solution

1. 1The n-th bright fringe lies at xₙ = nλD/d (D = screen distance, d = slit separation; neither is given in the question).
2. 2(a) For λ₁ = 650 nm and the third bright fringe: x₃ = 3λ₁D/d = 3 × 650 nm × D/d = 1950(D/d) nm.
3. 3(b) Bright fringes coincide when n₁λ₁ = n₂λ₂ ⇒ n₁/n₂ = 520/650 = 4/5. The least such pair is n₁ = 4, n₂ = 5.
4. 4Least distance x = n₂λ₂D/d = 5 × 520(D/d) nm = 2600(D/d) nm (equivalently 4 × 650(D/d) nm).