To solve the problem, we will use the formula for the position of the dark fringes in a Young's double-slit experiment. The formula for the position of the nth dark fringe is given by:
\[
x_n = \left( n - \frac{1}{2} \right) \frac{D \lambda}{d}
\]
Where:
- \( x_n \) is the distance of the nth dark fringe from the central maximum,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the distance between the two slits,
- \( \lambda \) is the wavelength of the light,
- \( n \) is the order of the dark fringe.
### Step 1: Identify the given values
- Distance between the two coherent sources (slits), \( d = 0.90 \, \text{mm} = 0.90 \times 10^{-3} \, \text{m} \)
- Distance from the slits to the screen, \( D = 1 \, \text{m} \)
- Distance of the second dark fringe from the central fringe, \( x_2 = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Order of the dark fringe, \( n = 2 \)
### Step 2: Substitute the values into the formula
Using the formula for the second dark fringe (\( n = 2 \)):
\[
x_2 = \left( 2 - \frac{1}{2} \right) \frac{D \lambda}{d}
\]
This simplifies to:
\[
x_2 = \frac{3}{2} \frac{D \lambda}{d}
\]
### Step 3: Rearrange the equation to solve for \( \lambda \)
Rearranging the equation to isolate \( \lambda \):
\[
\lambda = \frac{2 x_2 d}{3 D}
\]
### 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 numerator:
\[
2 \times (1 \times 10^{-3}) \times (0.90 \times 10^{-3}) = 1.8 \times 10^{-6} \, \text{m}
\]
Now, substituting this back into the equation for \( \lambda \):
\[
\lambda = \frac{1.8 \times 10^{-6}}{3} = 0.6 \times 10^{-6} \, \text{m} = 6 \times 10^{-7} \, \text{m}
\]
### Step 6: Convert to centimeters
Since the options are in centimeters, we convert:
\[
\lambda = 6 \times 10^{-7} \, \text{m} = 6 \times 10^{-5} \, \text{cm}
\]
### Final Answer
Thus, the wavelength of the monochromatic light used is:
\[
\lambda = 6 \times 10^{-5} \, \text{cm}
\]
