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Question: A plane wave front of wavelength 6000 Å is incident upon a slit of 0.2mm width, which enables Fraunh...

A plane wave front of wavelength 6000 Å is incident upon a slit of 0.2mm width, which enables Fraunhofer's diffraction pattern to be obtained on a screen 2 metre away. Width of the central maxima in mm will be :

A

10

B

8

C

12

D

2

Answer

12

Explanation

Solution

The problem involves calculating the width of the central maxima in a Fraunhofer single-slit diffraction pattern.

1. Identify Given Parameters:

  • Wavelength of light, λ=6000 A˚=6000×10−10 m=6×10−7 m\lambda = 6000 \text{ Å} = 6000 \times 10^{-10} \text{ m} = 6 \times 10^{-7} \text{ m}
  • Width of the slit, a=0.2 mm=0.2×10−3 m=2×10−4 ma = 0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m} = 2 \times 10^{-4} \text{ m}
  • Distance between the slit and the screen, D=2 metre=2 mD = 2 \text{ metre} = 2 \text{ m}

2. Formula for Width of Central Maxima: For a single-slit diffraction pattern, the angular position of the first minimum (on either side of the central maximum) is given by: asin⁡θ=nλa \sin\theta = n\lambda For the first minimum, n=1n=1, so asin⁡θ1=λa \sin\theta_1 = \lambda. For small angles (which is typically the case in Fraunhofer diffraction), sin⁡θ≈θ\sin\theta \approx \theta. Therefore, the angular position of the first minimum is θ1=λa\theta_1 = \frac{\lambda}{a}.

The linear distance of the first minimum from the central maximum on the screen, x1x_1, is given by: x1=Dtan⁡θ1≈Dθ1x_1 = D \tan\theta_1 \approx D \theta_1 (for small angles) Substituting θ1\theta_1: x1=Dλax_1 = \frac{D\lambda}{a}

The width of the central maxima (WW) is twice the distance of the first minimum from the central maximum (since the central maximum extends from the first minimum on one side to the first minimum on the other side). W=2x1=2DλaW = 2x_1 = \frac{2D\lambda}{a}

3. Calculation: Substitute the given values into the formula: W=2×(2 m)×(6×10−7 m)(2×10−4 m)W = \frac{2 \times (2 \text{ m}) \times (6 \times 10^{-7} \text{ m})}{(2 \times 10^{-4} \text{ m})} W=24×10−72×10−4W = \frac{24 \times 10^{-7}}{2 \times 10^{-4}} W=12×10−7×104W = 12 \times 10^{-7} \times 10^{4} W=12×10−3 mW = 12 \times 10^{-3} \text{ m}

4. Convert to Millimeters: To convert meters to millimeters, multiply by 1000 (1 m=1000 mm1 \text{ m} = 1000 \text{ mm}): W=12×10−3 m×1000 mm1 mW = 12 \times 10^{-3} \text{ m} \times \frac{1000 \text{ mm}}{1 \text{ m}} W=12×10−3×103 mmW = 12 \times 10^{-3} \times 10^{3} \text{ mm} W=12 mmW = 12 \text{ mm}

The width of the central maxima is 12 mm.