In Young's interference experiment in a large ripple tank, the vibrating sources are in phase and 120mm apart, the distance between adjacent maxima measured 2.00m away is 180mm. If the speed of the ripples is 54 cm/s, calculate the frequency of the vibrating source.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
- Distance between the sources (D) = 120 mm = 0.12 m
- Distance from the sources to the screen (L) = 2.00 m
- Distance between adjacent maxima (y) = 180 mm = 0.18 m
- Speed of the ripples (v) = 54 cm/s = 0.54 m/s

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