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Waves - Diffraction

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Diffraction is defined as the spreading out of waves when they pass through a gap (aperture) or travel around an obstacle.

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The extent of diffraction depends on the relationship between the wavelength λ\lambda and the width of the gap ww.

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Maximum diffraction occurs when the size of the gap is approximately equal to the wavelength of the wave, expressed as w≈λw \approx \lambda.

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If the gap width ww is much larger than the wavelength (w≫λw \gg \lambda), the waves pass through with very little spreading, resulting in a 'shadow' zone.

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Longer wavelengths (lower frequencies) diffract more than shorter wavelengths (higher frequencies) when passing through the same opening.

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Diffraction is a wave property; during diffraction, the wave speed vv, frequency ff, and wavelength λ\lambda remain constant.

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Sound waves have wavelengths ranging from centimeters to meters, which is why they diffract easily through doorways. Light waves have much smaller wavelengths (approx. 4×10−7 m4 \times 10^{-7}\text{ m} to 7×10−7 m7 \times 10^{-7}\text{ m}), so they require very narrow slits to show visible diffraction.

📐Formulae

v=fλv = f \lambda

Condition for significant diffraction: w≈λ\text{Condition for significant diffraction: } w \approx \lambda

Degree of diffraction∝λw\text{Degree of diffraction} \propto \frac{\lambda}{w}

💡Examples

Problem 1:

A sound wave has a frequency of 170 Hz170\text{ Hz} and travels at a speed of 340 m/s340\text{ m/s}. It approaches a doorway of width 1.0 m1.0\text{ m}. Calculate the wavelength and determine if significant diffraction will occur.

Solution:

First, calculate the wavelength using the wave equation: λ=vf\lambda = \frac{v}{f} λ=340170\lambda = \frac{340}{170} λ=2.0 m\lambda = 2.0\text{ m} Given the gap width w=1.0 mw = 1.0\text{ m}, we compare λ\lambda and ww. Since λ\lambda (2.0 m2.0\text{ m}) is of a similar order of magnitude to ww (1.0 m1.0\text{ m}), significant diffraction will occur.

Explanation:

Because the wavelength is comparable to the size of the opening, the sound waves will spread out significantly as they pass through the doorway.

Problem 2:

Two waves, AA and BB, pass through the same gap. Wave AA has a wavelength of 0.5 cm0.5\text{ cm} and Wave BB has a wavelength of 5.0 cm5.0\text{ cm}. Which wave will diffract more?

Solution:

The degree of diffraction is proportional to λw\frac{\lambda}{w}. For a fixed gap width ww, the wave with the larger wavelength will diffract more. λB>λA\lambda_{B} > \lambda_{A} 5.0 cm>0.5 cm5.0\text{ cm} > 0.5\text{ cm} Therefore, Wave BB will undergo more diffraction.

Explanation:

Longer wavelengths spread out more than shorter wavelengths when passing through the same aperture.