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. `60 xx 10^(-4)` cm
B. `10 xx 10^(4)` cm
C. `10 xx 10^(-5)` cm
D. `6 xx 10^(-5)` cm

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given parameters
We are given:
- Distance between the two slits (d) = 0.90 mm = \(0.90 \times 10^{-3}\) m
- Distance from the slits to the screen (D) = 1 m
- Position of the second dark fringe (Y_n) = 1 mm = \(1 \times 10^{-3}\) m
- We need to find the wavelength (\(\lambda\)) of the monochromatic light used.

### Step 2: Use the formula for dark fringes in Young's double-slit experiment
The position of the nth dark fringe in Young's experiment is given by the formula:
\[ Y_n = \frac{(2n - 1) \lambda D}{2d} \]
For the second dark fringe, \(n = 2\).

### Step 3: Substitute the known values into the formula
Substituting \(n = 2\), \(Y_n = 1 \times 10^{-3}\) m, \(D = 1\) m, and \(d = 0.90 \times 10^{-3}\) m into the formula:
\[ 1 \times 10^{-3} = \frac{(2 \cdot 2 - 1) \lambda \cdot 1}{2 \cdot (0.90 \times 10^{-3})} \]
This simplifies to:
\[ 1 \times 10^{-3} = \frac{3 \lambda}{2 \cdot 0.90 \times 10^{-3}} \]

### Step 4: Rearrange the equation to solve for \(\lambda\)
Rearranging the equation gives:
\[ \lambda = \frac{1 \times 10^{-3} \cdot 2 \cdot 0.90 \times 10^{-3}}{3} \]

### Step 5: Calculate \(\lambda\)
Calculating the right-hand side:
\[ \lambda = \frac{1 \times 10^{-3} \cdot 1.80 \times 10^{-3}}{3} \]
\[ \lambda = \frac{1.80 \times 10^{-6}}{3} = 0.60 \times 10^{-6} \text{ m} \]

### Step 6: Convert \(\lambda\) to a more convenient unit
Convert meters to centimeters:
\[ \lambda = 0.60 \times 10^{-6} \text{ m} = 6.0 \times 10^{-7} \text{ m} = 6.0 \times 10^{-5} \text{ cm} \]

### Final Answer
The wavelength of the monochromatic light used is \(6.0 \times 10^{-7}\) m or \(6.0 \times 10^{-5}\) cm.

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