In a Young's double slit experiment, the slit separation is `1mm` and the screen is `1m` from the slit. For a monochromatic light of wavelength `500nm`, the distance of 3rd minima from the central maxima is

Options

A. 0.50 mm
B. 1.25 mm
C. 1.50 mm
D. 1.75 mm

Answer

To find the distance of the 3rd minima from the central maxima in a Young's double slit experiment, we can use the formula for the position of minima, which is given by:

\[ X_n = \frac{(2n - 1) \lambda D}{2d} \]

where:
- \( X_n \) = distance of the nth minima from the central maxima
- \( n \) = order of the minima (for the 3rd minima, \( n = 3 \))
- \( \lambda \) = wavelength of the light
- \( D \) = distance from the slits to the screen
- \( d \) = slit separation

### Given Data:
- Slit separation, \( d = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \)
- Distance to the screen, \( D = 1 \text{ m} \)
- Wavelength, \( \lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} \)

### Step 1: Substitute the values into the formula
We need to find \( X_3 \) for the 3rd minima.

\[ X_3 = \frac{(2 \cdot 3 - 1) \cdot \lambda \cdot D}{2d} \]

### Step 2: Calculate \( (2 \cdot 3 - 1) \)
\[ (2 \cdot 3 - 1) = 6 - 1 = 5 \]

### Step 3: Substitute the values into the formula
Now substituting all the values into the equation:

\[ X_3 = \frac{5 \cdot (500 \times 10^{-9}) \cdot 1}{2 \cdot (1 \times 10^{-3})} \]

### Step 4: Simplify the expression
Calculating the denominator:

\[ 2 \cdot (1 \times 10^{-3}) = 2 \times 10^{-3} \]

Now substituting back:

\[ X_3 = \frac{5 \cdot 500 \times 10^{-9}}{2 \times 10^{-3}} \]

### Step 5: Calculate the numerator
\[ 5 \cdot 500 = 2500 \]
Thus,

\[ X_3 = \frac{2500 \times 10^{-9}}{2 \times 10^{-3}} \]

### Step 6: Divide the values
\[ X_3 = \frac{2500}{2} \times \frac{10^{-9}}{10^{-3}} = 1250 \times 10^{-6} \text{ m} \]

### Step 7: Convert to millimeters
\[ X_3 = 1.25 \times 10^{-3} \text{ m} = 1.25 \text{ mm} \]

### Final Answer
The distance of the 3rd minima from the central maxima is \( 1.25 \text{ mm} \).