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 \(2\times10^{-2}\,Wm^{-2}\).