In a Young's double slit experiment, the slits have unequal widths such that the ratio of the maximum intensity to the minimum intensity in the interference pattern is 25:9 . At a point P on the screen, the path difference between the interfering waves is 3 , where is the wavelength of the light used. If the intensity at point P is N times the intensity produced by the weaker slit alone, then the value of N is ______

Step-by-step solution

Let the intensities of the two slits be I₁ and I₂ , with I₁ > I₂ . The ratio of maximum to minimum intensity is given by: I_ max I_ min = ( I₁ + I₂ )^2 ( I₁ - I₂ )^2 = 25 9 Taking the square root of both sides: I₁ + I₂ I₁ - I₂ = 5 3 Cross-multiplying yields: 3 I₁ + 3 I₂ = 5 I₁ - 5 I₂ 8 I₂ = 2 I₁ I₁ = 4 I₂ I₁ = 16I₂ Let the intensity of the weaker slit be I₀ . Then I₂ = I₀ and I₁ = 16I₀ . The path difference at point P is x = 3 . The corresponding phase difference is: = 2 x = 2 ( 3 ) = 2 3 The resultant intensity I_