Question

In a YDSE setup slit separation is \(d = 5\lambda\), where \(\lambda\) is wavelength of light and screen is placed at distance \(D = 10d\). If maximum intensity is \(I_0\), find intensity at a point directly in front of one of the slit.

• A. \(I_0\)
• B. \(\frac{I_0}{3}\)
• C. \(\frac{I_0}{2}\)
• D. \(\frac{I_0}{6}\)

Hint:

In YDSE, \[ I = I_0 \cos^2\left(\frac{\Delta\phi}{2}\right) \] If path difference \(= \frac{\lambda}{4}\), phase difference \(= \frac{\pi}{2}\) and intensity becomes \(I_0/2\).

Answer 1

The Correct Option is C

Concept: In Young's Double Slit Experiment, \[ \Delta x = d \sin\theta \] Phase difference: \[ \Delta\phi = \frac{2\pi}{\lambda}\Delta x \] Intensity distribution: \[ I = I_0 \cos^2\left(\frac{\Delta\phi}{2}\right) \] where \(I_0\) is maximum intensity.
Step 1: Determine the path difference. Point is directly in front of slit \(S_1\). Vertical displacement from centre \[ y = \frac{d}{2} \] Given \[ D = 10d \] \[ \sin\theta = \frac{y}{\sqrt{y^2 + D^2}} \] \[ \sin\theta = \frac{d/2}{\sqrt{(d/2)^2 + (10d)^2}} \] \[ \sin\theta = \frac{d/2}{d\sqrt{1/4 + 100}} \] \[ \sin\theta = \frac{1}{2\sqrt{100.25}} \] Approximating, \[ \sin\theta \approx \frac{1}{20} \]
Step 2: Calculate path difference. \[ \Delta x = d\sin\theta \] \[ \Delta x = d \times \frac{1}{20} \] Given \(d = 5\lambda\) \[ \Delta x = \frac{5\lambda}{20} \] \[ \Delta x = \frac{\lambda}{4} \]
Step 3: Find phase difference. \[ \Delta\phi = \frac{2\pi}{\lambda}\Delta x \] \[ \Delta\phi = \frac{2\pi}{\lambda}\times\frac{\lambda}{4} \] \[ \Delta\phi = \frac{\pi}{2} \]
Step 4: Find intensity. \[ I = I_0\cos^2\left(\frac{\Delta\phi}{2}\right) \] \[ I = I_0\cos^2\left(\frac{\pi}{4}\right) \] \[ I = I_0 \left(\frac{1}{\sqrt2}\right)^2 \] \[ I = \frac{I_0}{2} \] \[ \boxed{I = \frac{I_0}{2}} \]