In certain Young's double slit experiment, the slit separation is `0.05` cm. The slit to screen distance is `100` cm. When blue light is used, the distance from central fringe to the fourth order bright fringe is `0.36` cm. what is thr wavelength of blue light

Options

A. `4000Å`
B. `4300Å`
C. `4400Å`
D. `4500Å`

Explanation

To find the wavelength of blue light used in the Young's double slit experiment, we can use the formula for the position of bright fringes in the interference pattern. The formula is given by:

\[ x = \frac{n \lambda D}{d} \]

Where:
- \( x \) = distance from the central fringe to the nth order bright fringe
- \( n \) = order of the bright fringe (in this case, \( n = 4 \))
- \( \lambda \) = wavelength of light
- \( D \) = distance from the slits to the screen
- \( d \) = slit separation

### Step-by-Step Solution:

1. **Identify the given values**:
- Slit separation, \( d = 0.05 \) cm = \( 0.05 \times 10^{-2} \) m = \( 5 \times 10^{-4} \) m
- Distance from slits to screen, \( D = 100 \) cm = \( 1 \) m
- Distance from central fringe to fourth-order bright fringe, \( x = 0.36 \) cm = \( 0.36 \times 10^{-2} \) m = \( 3.6 \times 10^{-3} \) m
- Order of the bright fringe, \( n = 4 \)

2. **Rearrange the formula to solve for \( \lambda \)**:
\[ \lambda = \frac{x \cdot d}{n \cdot D} \]

3. **Substitute the known values into the equation**:
\[ \lambda = \frac{(3.6 \times 10^{-3} \, \text{m}) \cdot (5 \times 10^{-4} \, \text{m})}{4 \cdot (1 \, \text{m})} \]

4. **Calculate the numerator**:
\[ 3.6 \times 10^{-3} \times 5 \times 10^{-4} = 1.8 \times 10^{-6} \, \text{m}^2 \]

5. **Calculate the denominator**:
\[ 4 \cdot 1 = 4 \]

6. **Now, calculate \( \lambda \)**:
\[ \lambda = \frac{1.8 \times 10^{-6}}{4} = 4.5 \times 10^{-7} \, \text{m} \]

7. **Convert the wavelength to nanometers**:
\[ \lambda = 4.5 \times 10^{-7} \, \text{m} = 4500 \, \text{nm} \]

### Final Answer:
The wavelength of blue light is \( 4500 \, \text{nm} \) or \( 4500 \, \text{Å} \).