As shown in the figure in YDSE experiment, if intensity is \( \frac{3}{4} \) of maximum intensity at point P, and path difference at point 'P' is \( \Delta x = \frac{\lambda}{\alpha} \). Find the value of "α". (where \( \lambda \) is wavelength of light)
Show Hint
In YDSE, the intensity at a point is related to the cosine square of the path difference. Use this relation to calculate the path difference and determine the corresponding value of \( \alpha \).
Step 1: Formula for intensity in YDSE.
In Young's Double Slit Experiment (YDSE), the intensity \( I \) at any point is related to the maximum intensity \( I_{\text{max}} \) by the formula:
\[
I = I_{\text{max}} \cos^2 \left( \frac{\pi \Delta x}{\lambda} \right)
\]
where \( \Delta x \) is the path difference and \( \lambda \) is the wavelength of light.
Step 2: Apply the given condition.
The intensity at point P is given as \( \frac{3}{4} \) of the maximum intensity:
\[
\frac{3}{4} I_{\text{max}} = I_{\text{max}} \cos^2 \left( \frac{\pi \Delta x}{\lambda} \right)
\]
Canceling \( I_{\text{max}} \) from both sides:
\[
\frac{3}{4} = \cos^2 \left( \frac{\pi \Delta x}{\lambda} \right)
\]
Taking the square root of both sides:
\[
\frac{\sqrt{3}}{2} = \cos \left( \frac{\pi \Delta x}{\lambda} \right)
\]
Step 3: Solve for \( \Delta x \).
The angle whose cosine is \( \frac{\sqrt{3}}{2} \) is \( \frac{\pi}{6} \):
\[
\frac{\pi \Delta x}{\lambda} = \frac{\pi}{6}
\]
Thus, the path difference \( \Delta x \) is:
\[
\Delta x = \frac{\lambda}{6}
\]
Step 4: Use the given path difference relation.
We are given that the path difference \( \Delta x = \frac{\lambda}{\alpha} \). From this, we have:
\[
\frac{\lambda}{6} = \frac{\lambda}{\alpha}
\]
Therefore, \( \alpha = 6 \).
Final Answer: 6
