Question

In the Young’s double slit experiment, the intensity of light passing through each of the two double slits is 2 × 10^-2 Wm^-2. The screen-slit distance is very large in comparison with slit-slit distance. The fringe width is β. The distance between the central maximum and a point P on the screen is \(\frac{\beta}{3}\). Then the total light intensity at that point is

• A. 4 × 10^-2 Wm^-2
• B. 2 × 10^-2 Wm^-2
• C. 16 × 10^-2 Wm^-2
• D. 8 × 10^-2 Wm^-2

Answer 1

The Correct Option is B

Given Information: 
Intensity of each slit, \(I = 2 \times 10^{-2}\, Wm^{-2}\)
Distance of the point from the central maximum, \(y = \frac{\beta}{3}\), where \(\beta\) is the fringe width.

Step-by-Step Explanation:

Step 1: Formula for Intensity in Young’s double-slit experiment:

The resultant intensity (\(I_{total}\)) at a point on the screen is given by:

\[ I_{total} = 4I\cos^2\left(\frac{\phi}{2}\right) \]

where \(I\) is the intensity from each slit, and \(\phi\) is the phase difference between waves from the two slits.

Step 2: Calculate the phase difference (\(\phi\)) at the given point clearly:

The distance between central maximum and the given point P is \(\frac{\beta}{3}\). The phase difference corresponding to one fringe width (\(\beta\)) is \(2\pi\). Thus, clearly:

\[ \phi = \frac{2\pi}{\beta}\times\frac{\beta}{3} = \frac{2\pi}{3} \]

Step 3: Substitute \(\phi = \frac{2\pi}{3}\) into intensity formula clearly:

\[ I_{total} = 4I\cos^2\left(\frac{\phi}{2}\right) = 4I\cos^2\left(\frac{\pi}{3}\right) \]

Since \(\cos\frac{\pi}{3} = \frac{1}{2}\):

\[ I_{total} = 4I\times\left(\frac{1}{2}\right)^2 = 4I\times\frac{1}{4} = I \]

Step 4: Clearly state the final intensity value:

\[ I_{total} = I = 2\times10^{-2}\,Wm^{-2} \]

Final Conclusion:
The total intensity at point P (at \(\frac{\beta}{3}\)) is clearly \(2\times10^{-2}\,Wm^{-2}\).

Answer 2

In Young's double slit experiment, the intensity of the interference pattern is given by: \[ I = I_0 \cos^2 \left( \frac{\pi x}{\beta} \right) \] Where \( I_0 \) is the maximum intensity (when the light from both slits is in phase), \( x \) is the distance from the central maximum to the point of observation, and \( \beta \) is the fringe width. Here, the intensity of light passing through each slit is given as \( I_0 = 2 \times 10^{-2} \, \text{W/m}^2 \), and the distance \( x \) from the central maximum to point \( P \) is \( \frac{\beta}{3} \). Now, substituting \( x = \frac{\beta}{3} \) into the intensity formula: \[ I = 2 \times 10^{-2} \, \cos^2 \left( \frac{\pi \cdot \frac{\beta}{3}}{\beta} \right) \] Simplifying: \[ I = 2 \times 10^{-2} \, \cos^2 \left( \frac{\pi}{3} \right) \] Since \( \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} \), we have: \[ I = 2 \times 10^{-2} \times \left( \frac{1}{2} \right)^2 = 2 \times 10^{-2} \times \frac{1}{4} = 2 \times 10^{-2} \times \frac{1}{4} = 2 \times 10^{-2} \, \text{W/m}^2 \] Thus, the total light intensity at point \( P \) is \( 2 \times 10^{-2} \, \text{W/m}^2 \).