In Young's double slit experiment the distance between slits is `2 xx 10^(-3)m`, the distance between screen and slits is 200 cm. When the light of wave length `5000 A^(@)` is used, the central maximum is at x=0 . the third maximum will be at x equal to

Options

A. 2cm
B. 0.05cm
C. 15cm
D. 0.5cm

Answer

To solve the problem, we will use the formula for the position of the m-th maximum in Young's double slit experiment, which is given by:

\[ x_m = \frac{m \lambda D}{d} \]

Where:
- \( x_m \) is the position of the m-th maximum,
- \( m \) is the order of the maximum (for the third maximum, \( m = 3 \)),
- \( \lambda \) is the wavelength of light,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the distance between the slits.

### Step 1: Identify the given values
- Distance between the slits, \( d = 2 \times 10^{-3} \, \text{m} \)
- Distance from the slits to the screen, \( D = 200 \, \text{cm} = 2 \, \text{m} \)
- Wavelength of light, \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} \)

### Step 2: Substitute the values into the formula
We need to find the position of the third maximum (\( m = 3 \)):
\[ x_3 = \frac{3 \lambda D}{d} \]
Substituting the values:
\[ x_3 = \frac{3 \times (5000 \times 10^{-10}) \times 2}{2 \times 10^{-3}} \]

### Step 3: Simplify the equation
Calculating the numerator:
\[ 3 \times (5000 \times 10^{-10}) \times 2 = 30000 \times 10^{-10} = 3 \times 10^{-6} \, \text{m} \]

Now substituting into the equation:
\[ x_3 = \frac{3 \times 10^{-6}}{2 \times 10^{-3}} = \frac{3}{2} \times 10^{-3} \, \text{m} = 1.5 \times 10^{-3} \, \text{m} \]

### Step 4: Convert to centimeters
To convert meters to centimeters:
\[ 1.5 \times 10^{-3} \, \text{m} = 1.5 \, \text{cm} \]

### Final Answer
The position of the third maximum is:
\[ x = 1.5 \, \text{cm} \]