A narrow slit of width 2 mm is illuminated by monochromatic light of wavelength 500nm. The distance between the first minima on either side on a screen at a distance of 1 m is

Options

A. 5 mm
B. 0.5 mm
C. 1 mm
D. 10 mm

Explanation

To find the distance between the first minima on either side of the central maximum on a screen, we can use the formula for single-slit diffraction. The position of the minima in a single-slit diffraction pattern is given by:

\[ y_m = \frac{m \lambda L}{a} \]

where:
- \(y_m\) is the position of the m-th minima from the central maximum,
- \(m\) is the order of the minima (for the first minima, \(m = 1\)),
- \(\lambda\) is the wavelength of the light,
- \(L\) is the distance from the slit to the screen,
- \(a\) is the width of the slit.

### Step 1: Identify the given values
- Width of the slit, \(a = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m}\)
- Wavelength of light, \(\lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m}\)
- Distance from the slit to the screen, \(L = 1 \, \text{m}\)

### Step 2: Calculate the position of the first minima
Using the formula for the first minima (\(m = 1\)):

\[ y_1 = \frac{1 \cdot (500 \times 10^{-9}) \cdot 1}{2 \times 10^{-3}} \]

Calculating this gives:

\[ y_1 = \frac{500 \times 10^{-9}}{2 \times 10^{-3}} = \frac{500}{2} \times 10^{-6} = 250 \times 10^{-6} \, \text{m} = 250 \, \mu\text{m} \]

### Step 3: Calculate the total distance between the first minima on either side
Since there is a minima on both sides of the central maximum, the total distance between the first minima on either side is:

\[ \text{Total distance} = 2y_1 = 2 \times 250 \, \mu\text{m} = 500 \, \mu\text{m} \]

### Final Answer
The distance between the first minima on either side on the screen is **500 μm**.

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