In a Young double slit experiment, the wavelength of incident light is 6000 Å, the separation between slits S₁ and S₂ is 5 cm and the distance between slits plane and screen is 50 cm, as shown in the figure below. If the resultant intensity at P is equal to the intensity due to individual slits, the path difference between interfering waves is __________ Å.

Step-by-step solution

Let the intensity due to each individual slit be I₀ . The resultant intensity I_R at a point where the phase difference is is given by: I_R = I₁ + I₂ + 2 I₁ I₂ Given that I₁ = I₂ = I₀ and the resultant intensity I_R = I₀ , we can substitute these values into the equation: I₀ = I₀ + I₀ + 2 I₀ I₀ I₀ = 2I₀ + 2I₀ I₀ = 2I₀ (1 + ) 1 + = 1 2 = - 1 2 The minimum phase difference satisfying this condition is: = 2 3 The relationship between phase difference and path difference x is: = 2 x Substituting the value of : 2 3 = 2