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 light at any point on the screen in a Young's double-slit experiment is determined by the interference of the two light waves reaching that point. The resultant intensity, denoted by \( I \), is calculated using the formula:

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

Where:

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

Assuming both waves have equal intensity \( I_0 \), so \( I_1 = I_2 = I_0 \), the formula for the resultant intensity simplifies to:

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

2. Phase Difference for a Path Difference of \( \frac{\lambda}{2} \):

The phase difference \( \delta \) and the path difference \( \Delta \) are related by the equation:

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

With a given path difference \( \Delta = \frac{\lambda}{2} \), the phase difference is calculated as:

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

3. Calculating Intensity at the Point:

Substituting \( \delta = \pi \) into the resultant intensity expression yields:

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

Since \( \cos \pi = -1 \), the intensity becomes:

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

4. Conclusion:

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