In a Young's experiment, two coherent sources are placed `0.90mm` apart and the fringes are observed one metre away. If is produces the second dark fringe at a distance of `1mm` from the central fringe, the wavelength of monochromatic light used would be

Options

A. (a) `60xx10^-4cm`
B. (b) `10xx10^-4cm`
C. (c) `10xx10^-5cm`
D. (d) `6xx10^-5cm`

Explanation

To solve the problem, we will use the formula for the position of dark fringes in Young's double-slit experiment. The position of the nth dark fringe is given by:

\[ X_n = \left(n - \frac{1}{2}\right) \frac{\lambda D}{d} \]

Where:
- \(X_n\) = position of the nth dark fringe from the central maximum
- \(n\) = order of the dark fringe (for the second dark fringe, \(n = 2\))
- \(\lambda\) = wavelength of the light used
- \(D\) = distance from the slits to the screen
- \(d\) = distance between the slits

### Step 1: Identify the known values
From the problem statement:
- Distance between the slits, \(d = 0.90 \, \text{mm} = 0.90 \times 10^{-3} \, \text{m}\)
- Distance from the slits to the screen, \(D = 1 \, \text{m}\)
- Position of the second dark fringe from the central maximum, \(X_2 = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m}\)

### Step 2: Substitute values into the dark fringe formula
Using the formula for the second dark fringe (\(n = 2\)):
\[ X_2 = \left(2 - \frac{1}{2}\right) \frac{\lambda D}{d} \]
This simplifies to:
\[ X_2 = \frac{3}{2} \frac{\lambda D}{d} \]

### Step 3: Rearranging the equation to solve for \(\lambda\)
We can rearrange the equation to find \(\lambda\):
\[ \lambda = \frac{2 X_2 d}{3 D} \]

### Step 4: Substitute the known values into the equation
Substituting the values we have:
\[ \lambda = \frac{2 \times (1 \times 10^{-3}) \times (0.90 \times 10^{-3})}{3 \times 1} \]

### Step 5: Calculate \(\lambda\)
Calculating the above expression:
\[ \lambda = \frac{2 \times 1 \times 0.90 \times 10^{-6}}{3} \]
\[ \lambda = \frac{1.8 \times 10^{-6}}{3} \]
\[ \lambda = 0.6 \times 10^{-6} \, \text{m} = 6 \times 10^{-7} \, \text{m} \]

### Final Answer
The wavelength of the monochromatic light used is:
\[ \lambda = 6 \times 10^{-7} \, \text{m} = 600 \, \text{nm} \]