In a Young double slit experiment, the wavelength of incident light is \(6000\,\text{\AA}\), the separation between slits \(S_1\) and \(S_2\) is \(5\,\text{cm}\) and the distance between slits plane and screen is \(50\,\text{cm}\), as shown in the figure. If the resultant intensity at \(P\) is equal to the intensity due to individual slits, the path difference between interfering waves is _____ \AA.
Step 1: Understanding the Question:
The problem asks for the path difference between waves at a point P where the total intensity is equal to the intensity of one slit. Step 2: Key Formula or Approach:
1. Resultant intensity formula: \(I = 4I_0 \cos^2\left(\frac{\phi}{2}\right)\), where \(I_0\) is the intensity of one slit.
2. Relation between phase difference (\(\phi\)) and path difference (\(\Delta x\)): \(\phi = \frac{2\pi}{\lambda} \Delta x\). Step 3: Detailed Explanation:
1. Given \(I = I_0\).
\[ I_0 = 4I_0 \cos^2\left(\frac{\phi}{2}\right) \Rightarrow \cos^2\left(\frac{\phi}{2}\right) = \frac{1}{4} \]
\[ \cos\left(\frac{\phi}{2}\right) = \frac{1}{2} \Rightarrow \frac{\phi}{2} = \frac{\pi}{3} \Rightarrow \phi = \frac{2\pi}{3} \]
2. Equating this to the path difference expression:
\[ \frac{2\pi}{3} = \frac{2\pi}{\lambda} \Delta x \]
\[ \Delta x = \frac{\lambda}{3} \]
3. Given \(\lambda = 6000 \text{ \AA}\):
\[ \Delta x = \frac{6000}{3} = 2000 \text{ \AA} \] Step 4: Final Answer:
The path difference at point P is 2000 \(\text{\AA}\).
