In YDSE, `D = 1 m, d = 1 mm`, and `lambda = 5000 nm`. The distance of the 100th maxima from the central maxima is

Options

A. `(1)/(2) m`
B. `(sqrt(3))/(2)` m
C. `(1)/(sqrt(3))` m
D. does not exist

Explanation

To solve the problem, we will use the formula for the distance of the nth maxima in Young's Double Slit Experiment (YDSE):

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Distance from the slits to the screen (D) = 1 m
- Distance between the slits (d) = 1 mm = \(1 \times 10^{-3}\) m
- Wavelength of light (\(\lambda\)) = 5000 nm = \(5000 \times 10^{-9}\) m

2. **Determine the Maximum Number of Fringes (n_max):**
The maximum number of fringes can be calculated using the formula:
\[ n_{max} = \frac{D}{\lambda} \]
Substituting the values:
\[ n_{max} = \frac{1 \, \text{m}}{5000 \times 10^{-9} \, \text{m}} = \frac{1}{5 \times 10^{-6}} = 200000 \]
However, we need to ensure that we do not exceed the maximum fringe number, which is determined by the geometry of the setup. The maximum number of fringes is limited by the condition \(D \sin \theta = n \lambda\).

3. **Calculate the Distance of the 100th Maxima (y_n):**
The formula for the distance of the nth maxima from the central maxima is given by:
\[ y_n = \frac{n \cdot D \cdot \lambda}{d} \]
For the 100th maxima (\(n = 100\)):
\[ y_{100} = \frac{100 \cdot 1 \, \text{m} \cdot 5000 \times 10^{-9} \, \text{m}}{1 \times 10^{-3} \, \text{m}} \]
Simplifying this:
\[ y_{100} = \frac{100 \cdot 5000 \times 10^{-9}}{10^{-3}} = 100 \cdot 5000 \times 10^{-6} = 0.5 \, \text{m} \]

4. **Final Result:**
Therefore, the distance of the 100th maxima from the central maxima is:
\[ y_{100} = 0.5 \, \text{m} \]