A Level Physics Practice Exam

If two coherent waves with a wavelength of 3.2 x 10^-2 m create a maximum at a detector, what is the path difference?

1. 1.6 x 10^-2 m

2. 6.4 x 10^-2 m

3. 3.2 x 10^-2 m

4. 1.28 m

The correct answer is: 3.2 x 10^-2 m

To determine the path difference that results in a maximum at a detector for two coherent waves, it's important to understand the principle of constructive interference. For constructive interference to occur, the path difference between the two waves must be an integer multiple of the wavelength. Given that the wavelength of the waves is \( 3.2 \times 10^{-2} \) m, the simplest scenario for constructive interference (creating a maximum) is when the path difference is equal to one full wavelength. This means that, for a maximum, the path difference can equal \( n \lambda \), where \( n \) is an integer (0, 1, 2, etc.) and \( \lambda \) is the wavelength. In the case of the first maximum (where \( n = 1 \)), the path difference will simply be \( \lambda \), which is \( 3.2 \times 10^{-2} \) m. Therefore, when the two waves meet at the detector with this path difference, they will interfere constructively, resulting in a maximum intensity at that point. This reasoning shows why the correct answer is the wavelength itself, confirming that \( 3.2 \times 10^{-2} \) m is