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Practice Section

Timed practice: Easy → Moderate → Advanced

⏱ Practice Timer

15:00

Recommended: 15 minutes for Easy, 20 for Moderate, 25 for Advanced

Easy Level Practice Problems

Problem 1

In YDSE, distance between slits d = 0.4 mm, distance to screen D = 1.5 m, wavelength λ = 600 nm. Calculate the fringe width β.

Answer:

β = λD/d = (600 × 10⁻⁹ × 1.5) / (0.4 × 10⁻³)

β = 2.25 × 10⁻³ m = 2.25 mm

Problem 2

State the conditions necessary for obtaining sustained interference of light.

Answer:

  1. Sources must be coherent (constant phase difference)
  2. Same wavelength/frequency
  3. Nearly equal amplitudes
  4. Sources should be close together
  5. Sources must be narrow

Problem 3

In single slit diffraction, if slit width is increased, what happens to the width of central maximum?

Answer:

Width decreases. Since width = 2λD/a, width is inversely proportional to slit width 'a'.

Problem 4

Light of intensity I₀ passes through a polaroid. What is the intensity of transmitted light?

Answer:

I = I₀/2 (if incident light is unpolarized)

Reason: Polaroid transmits only one component of electric field vector.

Problem 5

What is Brewster's angle for glass of refractive index 1.5?

Answer:

tan θ_B = μ = 1.5

θ_B = tan⁻¹(1.5) = 56.3°

Moderate Level Practice Problems

Problem 1

In YDSE with D = 1 m, d = 0.5 mm, a thin glass plate (μ = 1.5, thickness = 10 μm) is introduced in front of one slit. By how many fringes does the pattern shift? (λ = 500 nm)

Solution:

Additional path difference = (μ - 1)t = (1.5 - 1) × 10 × 10⁻⁶ = 5 × 10⁻⁶ m

Number of fringes shifted = Δx/λ = (5 × 10⁻⁶)/(500 × 10⁻⁹) = 10 fringes

Problem 2

Two slits separated by 0.2 mm are illuminated by light of wavelength 500 nm. If interference pattern is observed on screen 2 m away, find the position of 3rd dark fringe from center.

Solution:

Position of nth dark fringe: y = (2n - 1)λD/2d

For n = 3: y = (2×3 - 1) × 500 × 10⁻⁹ × 2 / (2 × 0.2 × 10⁻³)

y = 5 × 500 × 10⁻⁹ × 2 / (0.4 × 10⁻³) = 0.0125 m = 12.5 mm

Problem 3

In YDSE, intensities of two sources are in ratio 9:4. Find ratio of maximum to minimum intensity in interference pattern.

Solution:

Let I₁ = 9I and I₂ = 4I

I_max = (√I₁ + √I₂)² = (√(9I) + √(4I))² = (3√I + 2√I)² = 25I

I_min = (√I₁ - √I₂)² = (3√I - 2√I)² = I

Ratio = I_max/I_min = 25:1

Advanced Level Practice Problems

Problem 1

In YDSE, entire apparatus is immersed in liquid of refractive index n. If air is now slowly introduced into a slab of thickness t in front of one of the slits, how many fringes will cross a fixed point on the screen?

Solution:

Initial optical path in liquid = nt

Final optical path in air = t

Change in path difference = nt - t = (n - 1)t

Wavelength in liquid = λ/n

Number of fringes = Δx/λ_medium = (n - 1)t/(λ/n) = n(n - 1)t/λ

Problem 2

Show that average intensity over many fringes in an interference pattern equals I₁ + I₂, proving energy conservation.

Concept:

I = I₁ + I₂ + 2√(I₁I₂)cos δ

Average over one period: <I> = I₁ + I₂ + 2√(I₁I₂)<cos δ>

Since <cos δ> = 0 (averages to zero over complete cycle)

<I> = I₁ + I₂

Thus, energy redistributes spatially but total energy conserved.

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