A beam of light consisting of two wavelength , `6500 A^(@)` and `5200 A^(@)` is used to obtain interference fringes in a Young's double slit experiment `(1Å = 10^(-10)m)`. The distance between the slits 2.0 mm and the distance between the plane of the slits and the screen is 120 cm. Find the distance of the third bright fringe on the screen from the central maximum for the wavelength `6500 Å`.

Options

A. 0.117 cm
B. 0.28 cm
C. 0.05 cm
D. 0.3 cm

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert the Wavelength to Meters
The given wavelength is \( 6500 \, \text{Å} \). We need to convert this to meters.
\[ 6500 \, \text{Å} = 6500 \times 10^{-10} \, \text{m} = 6.5 \times 10^{-7} \, \text{m} \]

### Step 2: Identify the Given Values
- Distance between the slits (d): \( 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
- Distance from the slits to the screen (D): \( 120 \, \text{cm} = 1.2 \, \text{m} \)

### Step 3: Use the Formula for the Position of the Bright Fringe
The position of the \( n \)-th bright fringe in a Young's double slit experiment is given by:
\[ y_n = \frac{n \lambda D}{d} \]
For the third bright fringe (\( n = 3 \)):
\[ y_3 = \frac{3 \lambda D}{d} \]

### Step 4: Substitute the Values into the Formula
Substituting the values we have:
\[ y_3 = \frac{3 \times (6.5 \times 10^{-7} \, \text{m}) \times (1.2 \, \text{m})}{2 \times 10^{-3} \, \text{m}} \]

### Step 5: Calculate the Value
Calculating the above expression:
\[ y_3 = \frac{3 \times 6.5 \times 1.2}{2} \times 10^{-4} \, \text{m} \]
\[ y_3 = \frac{23.4}{2} \times 10^{-4} \, \text{m} = 11.7 \times 10^{-4} \, \text{m} = 0.00117 \, \text{m} \]

### Step 6: Convert to Centimeters
To convert to centimeters:
\[ y_3 = 0.00117 \, \text{m} \times 100 \, \text{cm/m} = 0.117 \, \text{cm} \]

### Final Answer
The distance of the third bright fringe from the central maximum for the wavelength \( 6500 \, \text{Å} \) is:
\[ \boxed{0.117 \, \text{cm}} \]

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