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Wave Optics - Huygens Principle

Grade 12CBSEPhysics

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

🔑Concepts

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A Wavefront is defined as the locus of all points in a medium that vibrate in the same phase at any given instant. For a point source, the wavefronts are spherical, while for a distant source, they are plane wavefronts.

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Huygens Principle states that every point on a primary wavefront acts as a source of secondary spherical wavelets. These wavelets spread out in all directions with the speed of the wave in that medium.

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The new position of the wavefront at a later time t+Δtt + \Delta t is the forward envelope (tangential surface) of all the secondary wavelets at that instant, having a radius r=vΔtr = v \Delta t.

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Using Huygens Principle, the Law of Reflection can be derived by showing that the angle of incidence ii is equal to the angle of reflection rr (i=ri = r).

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The Law of Refraction (Snell's Law) is verified by considering the change in speed of light as it moves from one medium to another, leading to the relation sin⁡isin⁡r=v1v2=n2n1\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \frac{n_2}{n_1}.

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When light travels from a rarer medium to a denser medium, its speed decreases (v<cv < c), and its wavelength decreases (lambda′<lambda\\lambda' < \\lambda), but the frequency ff remains constant.

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A plane wavefront passing through a convex lens converges into a spherical wavefront towards the focus, while passing through a concave lens results in a diverging spherical wavefront.

📐Formulae

n=cvn = \frac{c}{v}

sin⁡isin⁡r=v1v2=n2n1\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \frac{n_2}{n_1}

λmedium=λvacuumn\lambda_{medium} = \frac{\lambda_{vacuum}}{n}

v=fλv = f \lambda

Δϕ=2πλΔx\Delta \phi = \frac{2\pi}{\lambda} \Delta x

💡Examples

Problem 1:

Monochromatic light of wavelength 589 nm589\text{ nm} is incident from air on a water surface. If the refractive index of water is 1.331.33, calculate the wavelength and speed of light in water.

Solution:

Given λair=589 nm\lambda_{air} = 589\text{ nm} and n=1.33n = 1.33. The speed of light in vacuum c≈3×108 m/sc \approx 3 \times 10^8\text{ m/s}.

  1. Speed in water: v=cn=3×1081.33≈2.25×108 m/sv = \frac{c}{n} = \frac{3 \times 10^8}{1.33} \approx 2.25 \times 10^8\text{ m/s}.
  2. Wavelength in water: λwater=λairn=5891.33≈442.86 nm\lambda_{water} = \frac{\lambda_{air}}{n} = \frac{589}{1.33} \approx 442.86\text{ nm}.

Explanation:

The frequency remains unchanged when light enters a new medium, but the speed and wavelength decrease in proportion to the refractive index nn.

Problem 2:

Explain the shape of the wavefront when a plane wavefront is incident on a thin convex lens.

Solution:

When a plane wavefront is incident on a convex lens, the central part of the wavefront travels through the thickest part of the lens and is delayed the most. The outer parts travel through thinner sections and are delayed less. This results in the plane wavefront emerging as a converging spherical wavefront.

Explanation:

According to Huygens Principle, different parts of the wavefront experience different time delays based on the optical path length, causing the change in the geometry of the wavefront.

Huygens Principle - Revision Notes & Key Formulas | CBSE Class 12 Physics