Saurabh Koranglekar
Last Activity: 5 Years ago
To determine the distance between areas of total darkness on the screen when two different wavelengths interfere in a double-slit experiment, we look for the condition when the bright fringes of both wavelengths exactly overlap. These are known as coincidence minima or common minima.
Key Concept: Common Minima in Interference Pattern
In a double-slit setup, the condition for constructive interference (bright fringe) for a given wavelength λ is:
y = (nλD) / d
Where:
- y is the fringe position on the screen
- n is an integer (fringe order)
- λ is the wavelength of light
- D is the distance from the slit to the screen
- d is the slit separation
Total darkness (complete destructive interference for both wavelengths) will occur at positions where bright fringes of both wavelengths coincide again. This happens at the least common multiple (LCM) of the fringe spacings of both wavelengths.
Step 1: Given Values
- λ₁ = 400 nm = 400 × 10⁻⁹ m
- λ₂ = 560 nm = 560 × 10⁻⁹ m
- d = 0.1 mm = 1 × 10⁻⁴ m
- D = 1 m
Step 2: Fringe Spacing for Each Wavelength
y₁ = (λ₁ × D) / d = (400 × 10⁻⁹ × 1) / (1 × 10⁻⁴) = 4 × 10⁻³ m = 4 mm
y₂ = (λ₂ × D) / d = (560 × 10⁻⁹ × 1) / (1 × 10⁻⁴) = 5.6 × 10⁻³ m = 5.6 mm
Step 3: Find LCM of Fringe Spacings
We are looking for the smallest distance where both sets of bright fringes overlap again. This is simply the LCM of 4 mm and 5.6 mm.
LCM of 4 and 5.6 is:
LCM(4, 5.6) = LCM(40, 56) / 10 = 280 / 10 = 28 mm
Final Result
✅ The distance between areas of total darkness is 28 mm.
Correct Answer:
(D) 28 mm
