In Young's double slit experiment, 'I' is the minimum intensity and ' \(\mathrm{I}_1\) ' is the intensity at a point where the path difference is \(\frac{\lambda}{4}\) where ' \(\lambda\) ' is the wavelength of light used. The ratio \(I / I_1\) is (Intensities of the two interfering waves are same) \(\left(\cos 0^{\circ}=1, \cos 90^{\circ}=0\right)\)

- Concept: At \(\lambda / 4\) path difference, the phase difference is \(\pi / 2\) radians. The resultant intensity for two waves of equal amplitude interfering with a phase difference \(\phi\) is \(I=4 I_0 \cos ^2(\phi / 2)\).
- Calculation: \(\phi=\pi / 2\) leads to \(I=4 I_0 \cos ^2(\pi / 4)=4 I_0 \times(1 / \sqrt{2})^2=2 I_0\).
- Since \(I_1\) (dark fringe) is 0 , comparing intensities is not straightforward, but given that minimum intensity \(\left(I_1\right)\) typically represents a baseline or zero in the theoretical model, the scenario described doesn't make practical sense. It's a conceptual error as \(I_1\) cannot be zero in an actual setting where other fringes are being compared unless explicitly defined differently in a particular experimental setup.
Answer: The theoretical response is conceptually flawed because the minimum intensity in a typical Young's double slit setup (constructive and destructive interference model) is not zero, it's rather \(I_0\). Assuming \(I_1=I_0, I / I_1=2\).