In Young's experiment, the pinholes are illuminated by a monochromatic light of wavelengths `5200Å`. A screen is palaced at 1m. From the sli. If the pinhole are `1.3` mm apart, then distancebetween the sevength bright band on one side and sixth dark band on the other side of the central bright band will be

Options

A. `20xx10^(-4)m `
B. `50xx10^(-4)m `
C. `30xx10^(-4)m `
D. `60xx10^(-4)m `

Explanation

To solve the problem step by step, we will use the principles of Young's Double Slit Experiment (YDSE).

### Step 1: Identify the given values
- Wavelength of light, \( \lambda = 5200 \, \text{Å} = 5200 \times 10^{-10} \, \text{m} \)
- Distance from the slits to the screen, \( D = 1 \, \text{m} \)
- Distance between the pinholes, \( d = 1.3 \, \text{mm} = 1.3 \times 10^{-3} \, \text{m} \)

### Step 2: Calculate the position of the 7th bright band
The formula for the position of the nth bright fringe in YDSE is given by:
\[ y_n = \frac{n \lambda D}{d} \]
For the 7th bright band (\( n = 7 \)):
\[ y_7 = \frac{7 \cdot \lambda \cdot D}{d} \]

### Step 3: Substitute the values for the 7th bright band
Substituting the known values:
\[ y_7 = \frac{7 \cdot (5200 \times 10^{-10}) \cdot 1}{1.3 \times 10^{-3}} \]

### Step 4: Calculate \( y_7 \)
Calculating \( y_7 \):
\[ y_7 = \frac{36400 \times 10^{-10}}{1.3 \times 10^{-3}} = \frac{36400 \times 10^{-10}}{1.3 \times 10^{-3}} = \frac{36400}{1.3} \times 10^{-7} \approx 28000 \times 10^{-7} \, \text{m} = 0.0028 \, \text{m} \]

### Step 5: Calculate the position of the 6th dark band
The formula for the position of the nth dark fringe is given by:
\[ y_n = \frac{(2n - 1) \lambda D}{2d} \]
For the 6th dark band (\( n = 6 \)):
\[ y_6 = \frac{(2 \cdot 6 - 1) \lambda D}{2d} = \frac{11 \lambda D}{2d} \]

### Step 6: Substitute the values for the 6th dark band
Substituting the known values:
\[ y_6 = \frac{11 \cdot (5200 \times 10^{-10}) \cdot 1}{2 \cdot (1.3 \times 10^{-3})} \]

### Step 7: Calculate \( y_6 \)
Calculating \( y_6 \):
\[ y_6 = \frac{57200 \times 10^{-10}}{2.6 \times 10^{-3}} = \frac{57200}{2.6} \times 10^{-7} \approx 22000 \times 10^{-7} \, \text{m} = 0.0022 \, \text{m} \]

### Step 8: Calculate the total distance between the 7th bright band and the 6th dark band
The total distance \( Y \) between the 7th bright band and the 6th dark band is:
\[ Y = y_7 + y_6 \]
Substituting the values:
\[ Y = 0.0028 \, \text{m} + 0.0022 \, \text{m} = 0.005 \, \text{m} = 5 \, \text{mm} \]

### Final Answer
The distance between the 7th bright band on one side and the 6th dark band on the other side of the central bright band is **5 mm**.

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