In Young's double slit experiment, 'I' is the minimum intensity and ' $\mathrm{I}_1$ ' is the intensity at a…
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) $\left(\cos 0^{\circ}=1, \cos 90^{\circ}=0\right)$
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Solution
- 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 $I_1=I_0, I / I_1=2$.