Question

In a Young's double-slit experiment, two light waves, each of intensity \( I_0 \), interfere at a point, having a path difference \( \frac{\lambda}{2} \) on the screen. Find the intensity at this point.

Hint:

When the path difference between two waves is \( \frac{\lambda}{2} \), the waves are in destructive interference, and the intensity at that point will be zero.

Answer 1

1. Intensity and Interference: 

The intensity of the light at any point on the screen in a Young’s double-slit experiment depends on the interference of the two light waves reaching that point. The resultant intensity \( I \) is given by the formula:

\[ I = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos \delta \]

Where:

• \( I_1 \) and \( I_2 \) are the intensities of the individual waves.
• \( \delta \) is the phase difference between the two waves.

 

Since both waves have the same intensity \( I_0 \), we have \( I_1 = I_2 = I_0 \). So, the expression for the resultant intensity becomes:

\[ I = 2I_0 + 2I_0 \cos \delta \]

2. Phase Difference at Path Difference \( \frac{\lambda}{2} \):

The phase difference \( \delta \) is related to the path difference \( \Delta \) by the equation:

\[ \delta = \frac{2 \pi \Delta}{\lambda} \]

Given that the path difference \( \Delta = \frac{\lambda}{2} \), we substitute this into the equation for the phase difference:

\[ \delta = \frac{2 \pi \times \frac{\lambda}{2}}{\lambda} = \pi \]

3. Calculating the Intensity at the Point:

Now, substitute \( \delta = \pi \) into the expression for the resultant intensity:

\[ I = 2I_0 + 2I_0 \cos \pi \]

Since \( \cos \pi = -1 \), we get:

\[ I = 2I_0 + 2I_0 \times (-1) = 2I_0 - 2I_0 = 0 \]

4. Conclusion:

• The intensity at the point where the path difference is \( \frac{\lambda}{2} \) is \( I = 0 \).