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. (a) `0.50mm`
B. (b) `1.25mm`
C. (c) `1.50mm`
D. (d) `1.75mm`

Explanation

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

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

Where:
- \(x_n\) is the distance of the nth minima from the central maximum.
- \(n\) is the order of the minima (in this case, \(n = 3\)).
- \(\lambda\) is the wavelength of the light used.
- \(D\) is the distance from the slits to the screen.
- \(d\) is the separation between the slits.

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

### Step-by-Step Solution:

1. **Identify the values:**
- \(n = 3\)
- \(\lambda = 500 \times 10^{-9} \text{ m}\)
- \(D = 1 \text{ m}\)
- \(d = 1 \times 10^{-3} \text{ m}\)

2. **Substitute the values into the formula:**
\[ x_3 = \frac{(2 \times 3 - 1) \lambda D}{2d} \]
\[ x_3 = \frac{(6 - 1) \cdot (500 \times 10^{-9}) \cdot 1}{2 \cdot (1 \times 10^{-3})} \]

3. **Calculate the numerator:**
\[ x_3 = \frac{5 \cdot (500 \times 10^{-9}) \cdot 1}{2 \cdot (1 \times 10^{-3})} \]
\[ = \frac{2500 \times 10^{-9}}{2 \times 10^{-3}} \]

4. **Simplify the expression:**
\[ = \frac{2500 \times 10^{-9}}{2 \times 10^{-3}} = \frac{2500}{2} \times 10^{-9 + 3} \]
\[ = 1250 \times 10^{-6} \text{ m} \]

5. **Convert to millimeters:**
\[ = 1.25 \text{ mm} \]

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

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