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

Step 1: Understanding the Concept:
In Young's Double Slit Experiment (YDSE), the intensity at any point on the screen depends on the path difference between the waves reaching that point.
We need to find the path difference for a point exactly in front of one of the slits.
Step 2: Key Formula or Approach:
The path difference $\Delta x$ at a vertical distance $y$ is approximately $\Delta x = \frac{yd}{D}$.
The phase difference is $\Delta\phi = \frac{2\pi}{\lambda}\Delta x$.
The resultant intensity is $I = I_0 \cos^2 \left( \frac{\Delta\phi}{2} \right)$.
Step 3: Detailed Explanation:
The point directly in front of one slit is at a distance $y = \frac{d}{2}$ from the central maximum.
Given $D = 10d$ and $d = 5\lambda$.
The path difference $\Delta x$ is:
\[ \Delta x \approx \frac{yd}{D} = \frac{(d/2) \cdot d}{10d} = \frac{d}{20} \]
Substituting $d = 5\lambda$:
\[ \Delta x = \frac{5\lambda}{20} = \frac{\lambda}{4} \]
Now, compute the phase difference $\Delta\phi$:
\[ \Delta\phi = \frac{2\pi}{\lambda} \times \Delta x = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2} \]
Finally, the intensity $I$ at this point is:
\[ I = I_0 \cos^2\left(\frac{\Delta\phi}{2}\right) = I_0 \cos^2\left(\frac{\pi/2}{2}\right) = I_0 \cos^2\left(\frac{\pi}{4}\right) \]
\[ I = I_0 \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2} \]
Step 4: Final Answer:
The intensity directly in front of one of the slits is I$_0$/2.