Correct Option: (b) \(\frac{I_0}{2}\)
Step 1 (Formula / Governing Law & Given Data): [0.25 Mark]
The intensity \(I\) at any point on the screen in Young's Double Slit Experiment is given by the relation:
$$I = I_{max} \cos^2\left(\frac{\phi}{2}\right)$$Where \(\phi\) is the phase difference, related to path difference \(\Delta x\) by: \(\phi = \frac{2\pi}{\lambda} \Delta x\).
Step 2 (Step-by-Step Mathematical Working): [0.5 Mark]
- Case 1: Given path difference \(\Delta x_1 = \lambda\).
Phase difference \(\phi_1 = \frac{2\pi}{\lambda} \times \lambda = 2\pi\).
Intensity \(I_0 = I_{max} \cos^2\left(\frac{2\pi}{2}\right) = I_{max} \cos^2(\pi) = I_{max}(1)^2 = I_{max}\). - Case 2: Given path difference \(\Delta x_2 = \frac{\lambda}{4}\).
Phase difference \(\phi_2 = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}\).
Intensity \(I = I_{max} \cos^2\left(\frac{\pi/2}{2}\right) = I_{max} \cos^2\left(\frac{\pi}{4}\right)\).
Step 3 (Final Concluded Answer): [0.25 Mark]
Substituting the value of \(\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\):
$$I = I_{max} \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_{max}}{2}$$Since \(I_{max} = I_0\), the intensity is \(\frac{I_0}{2}\).
💡 EL10 Concept Insight: Think of light waves like two people jumping on a trampoline; when they are perfectly in sync (path difference \(\lambda\), they create a big splash, but when they are a quarter-step out of sync (path difference \(\lambda/4\)), they cancel each other out just enough to leave you with exactly half the total energy.