Light of wavelength `6.5xx10^(-7)`m is made incident on two slits 1 mm apart. The distance between third dark fringe and fifth bright fringe on a screen distant 1 m from the slits will be

Options

A. 0.325 mm
B. 0.65 mm
C. 1.625 mm
D. 3.25 mm

Explanation

To solve the problem of finding the distance between the third dark fringe and the fifth bright fringe in a double-slit interference pattern, we will follow these steps:

### Step 1: Understand the given data
- Wavelength of light, \( \lambda = 6.5 \times 10^{-7} \) m
- Distance between the slits, \( d = 1 \) mm = \( 1 \times 10^{-3} \) m
- Distance from the slits to the screen, \( D = 1 \) m

### Step 2: Determine the formula for the position of bright and dark fringes
1. The position of the \( n \)-th bright fringe is given by:
\[ y_b = \frac{n \lambda D}{d} \]
2. The position of the \( n \)-th dark fringe is given by:
\[ y_d = \frac{(2n-1) \lambda D}{2d} \]

### Step 3: Calculate the position of the 5th bright fringe
For \( n = 5 \):
\[ y_b = \frac{5 \lambda D}{d} = \frac{5 \times (6.5 \times 10^{-7}) \times 1}{1 \times 10^{-3}} \]
Calculating this:
\[ y_b = \frac{5 \times 6.5 \times 10^{-7}}{1 \times 10^{-3}} = 5 \times 6.5 \times 10^{-4} = 3.25 \times 10^{-3} \text{ m} \]

### Step 4: Calculate the position of the 3rd dark fringe
For \( n = 3 \):
\[ y_d = \frac{(2 \times 3 - 1) \lambda D}{2d} = \frac{5 \lambda D}{2d} = \frac{5 \times (6.5 \times 10^{-7}) \times 1}{2 \times 1 \times 10^{-3}} \]
Calculating this:
\[ y_d = \frac{5 \times 6.5 \times 10^{-7}}{2 \times 10^{-3}} = \frac{5 \times 6.5 \times 10^{-4}}{2} = 1.625 \times 10^{-3} \text{ m} \]

### Step 5: Calculate the distance between the 5th bright fringe and the 3rd dark fringe
Now, we find the distance between the 5th bright fringe and the 3rd dark fringe:
\[ \text{Distance} = y_b - y_d = (3.25 \times 10^{-3}) - (1.625 \times 10^{-3}) = 1.625 \times 10^{-3} \text{ m} \]

### Step 6: Convert the distance to mm
To convert meters to millimeters:
\[ 1.625 \times 10^{-3} \text{ m} = 1.625 \text{ mm} \]

### Final Answer
The distance between the third dark fringe and the fifth bright fringe is \( 1.625 \text{ mm} \).

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