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Calculate Angular Width of Central Maximum in Single Slit Diffraction

In single slit diffraction pattern, the wavelength of light used is 628 nm628\text{ nm} and slit width is 0.2 mm0.2\text{ mm}, the angular width of central maximum is α×10−2\alpha \times 10^{-2} degrees. The value of α\alpha is ________.

Official Numerical Answer36

Topics & Concepts

Wave OpticsDiffraction

Step-by-Step Solution

To find the angular width of the central maximum in a single-slit diffraction pattern, we use the diffraction formula for the first minima on either side of the central maximum.

1. Given Data:

  • Wavelength of light, λ=628 nm=628×10−9 m\lambda = 628\text{ nm} = 628 \times 10^{-9}\text{ m}
  • Slit width, 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}
  • Angular width of the central maximum =α×10−2 degrees= \alpha \times 10^{-2}\text{ degrees}

2. Formula for Angular Width:

The condition for the first minimum in a single-slit diffraction pattern is given by: asin⁡θ=λa \sin\theta = \lambda

For small angles θ\theta, we can approximate sin⁡θ≈θ\sin\theta \approx \theta (in radians): θ≈λa\theta \approx \frac{\lambda}{a}

The angular width of the central maximum (θ0\theta_0) is the angular separation between the first minima on both sides of the central maximum: θ0=2θ=2λa radians\theta_0 = 2\theta = \frac{2\lambda}{a}\text{ radians}

3. Calculation in Radians:

Substitute the given values into the equation: θ0=2×628×10−9 m0.2×10−3 m\theta_0 = \frac{2 \times 628 \times 10^{-9}\text{ m}}{0.2 \times 10^{-3}\text{ m}}

θ0=6280×10−6 rad=6.28×10−3 rad\theta_0 = 6280 \times 10^{-6}\text{ rad} = 6.28 \times 10^{-3}\text{ rad}

Since π≈3.14\pi \approx 3.14, we can write 6.286.28 as 2π2\pi: θ0=2π×10−3 rad\theta_0 = 2\pi \times 10^{-3}\text{ rad}

4. Conversion to Degrees:

To convert radians to degrees, we multiply by 180∘π\frac{180^\circ}{\pi}: θ0=(2π×10−3 rad)×180∘π\theta_0 = \left(2\pi \times 10^{-3}\text{ rad}\right) \times \frac{180^\circ}{\pi}

θ0=2×180×10−3∘=360×10−3∘=36×10−2∘\theta_0 = 2 \times 180 \times 10^{-3\circ} = 360 \times 10^{-3\circ} = 36 \times 10^{-2\circ}

5. Comparison:

Comparing our result with the given expression α×10−2 degrees\alpha \times 10^{-2}\text{ degrees}: α×10−2=36×10−2\alpha \times 10^{-2} = 36 \times 10^{-2}

α=36\alpha = 36

Calculate Angular Width of Central Maximum in Single Slit Diffraction | Physics PYQ Solution - JEE Challenger