In a double slit experiment , the slit separation is `0.20` cm and the slit to screen distance is 100 cm. The positions of the first three minima, if wavelength of the source is 500 nm is

Options

A. `+-0.125nm, +-0.375cm, +-0.625cm`
B. `+-0.025cm, +-0.075cm, +-0.125 cm`
C. `+-12.5cm, +-37.5cm, +-62.5cm`
D. `+-1.25cm, +- 3.75 cm, +- 6.25 cm`

Explanation

To solve the problem of finding the positions of the first three minima in a double slit experiment, we can follow these steps:

### Step 1: Understand the setup
In a double slit experiment, we have two slits separated by a distance \( d \) and a screen at a distance \( D \) from the slits. The wavelength of the light used is \( \lambda \).

Given:
- Slit separation, \( d = 0.20 \, \text{cm} = 0.20 \times 10^{-2} \, \text{m} \)
- Distance from slits to screen, \( D = 100 \, \text{cm} = 1.00 \, \text{m} \)
- Wavelength, \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \)

### Step 2: Formula for the position of minima
The position of the \( n \)-th minima in a double slit experiment is given by the formula:
\[ X_n = \frac{(2n - 1) \lambda D}{2d} \]
where \( n \) is the order of the minima (1 for the first minima, 2 for the second, etc.).

### Step 3: Calculate the first minima (\( n = 1 \))
Substituting \( n = 1 \):
\[ X_1 = \frac{(2 \cdot 1 - 1) \lambda D}{2d} = \frac{1 \cdot (500 \times 10^{-9}) \cdot (1)}{2 \cdot (0.20 \times 10^{-2})} \]
Calculating:
\[ X_1 = \frac{500 \times 10^{-9}}{0.40 \times 10^{-2}} = \frac{500 \times 10^{-9}}{4 \times 10^{-3}} = 125 \times 10^{-6} \, \text{m} = 0.125 \, \text{mm} \]

### Step 4: Calculate the second minima (\( n = 2 \))
Substituting \( n = 2 \):
\[ X_2 = \frac{(2 \cdot 2 - 1) \lambda D}{2d} = \frac{3 \cdot (500 \times 10^{-9}) \cdot (1)}{2 \cdot (0.20 \times 10^{-2})} \]
Calculating:
\[ X_2 = \frac{1500 \times 10^{-9}}{0.40 \times 10^{-2}} = \frac{1500 \times 10^{-9}}{4 \times 10^{-3}} = 375 \times 10^{-6} \, \text{m} = 0.375 \, \text{mm} \]

### Step 5: Calculate the third minima (\( n = 3 \))
Substituting \( n = 3 \):
\[ X_3 = \frac{(2 \cdot 3 - 1) \lambda D}{2d} = \frac{5 \cdot (500 \times 10^{-9}) \cdot (1)}{2 \cdot (0.20 \times 10^{-2})} \]
Calculating:
\[ X_3 = \frac{2500 \times 10^{-9}}{0.40 \times 10^{-2}} = \frac{2500 \times 10^{-9}}{4 \times 10^{-3}} = 625 \times 10^{-6} \, \text{m} = 0.625 \, \text{mm} \]

### Final Results
The positions of the first three minima are:
- First minima: \( 0.125 \, \text{mm} \)
- Second minima: \( 0.375 \, \text{mm} \)
- Third minima: \( 0.625 \, \text{mm} \)