In the youngs double slit experiment ,the slit sepration is 1mm ,the distance between source and screen is 1m ,wavelength of light used is 6500 angstrom.Find the distance between fifth maxima and third minima in (mm)?

Explanation

To solve the problem of finding the distance between the fifth maxima and the third minima in Young's double slit experiment, we will follow these steps:

### Step 1: Write down the given data
- Slit separation (d) = 1 mm = \(1 \times 10^{-3}\) m
- Distance between source and screen (D) = 1 m
- Wavelength of light (λ) = 6500 Å = \(6500 \times 10^{-10}\) m = \(6.5 \times 10^{-7}\) m

### Step 2: Calculate the position of the fifth maxima
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