In Young's double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths λ₁=12000Å and λ₂=10000Å. At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

Let λ₁ = 12000 Å = 12000 × 10⁻¹⁰ m and λ₂ = 10000 Å = 10000 × 10⁻¹⁰ m.
The distance between the slits, d = 2 mm = 2 × 10⁻³ m.
The distance between the slits and the screen, D = 2 m.
For bright fringes, the condition is given by:
x = nλD/d, where n is the order of the bright fringe.
Let x be the distance from the central bright fringe where a bright fringe from one interference pattern coincides with a bright fringe from the other.
Then, for λ₁, x = n₁λ₁D/d
For λ₂, x = n₂λ₂D/d
Since the bright fringes coincide, x is the same for both wavelengths. Therefore,
n₁λ₁D/d = n₂λ₂D/d
n₁λ₁ = n₂λ₂
n₁/n₂ = λ₂/λ₁ = 10000/12000 = 5/6
The minimum values of n₁ and n₂ that satisfy this ratio are n₁ = 5 and n₂ = 6.
Substituting n₁ = 5 and λ₁ = 12000 × 10⁻¹⁰ m into the equation for x:
x = n₁λ₁D/d = 5 × (12000 × 10⁻¹⁰ m) × (2 m) / (2 × 10⁻³ m)
x = 6 × 10⁻³ m = 6 mm
Therefore, the minimum distance from the common central bright fringe where a bright fringe from one interference pattern coincides with a bright fringe from the other is 6 mm.