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Wave Optics - Refraction and Reflection of Plane Waves using Huygens Principle

Grade 12CBSEPhysics

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

🔑Concepts

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Huygens Principle: Every point on a given wavefront acts as a source of secondary spherical wavelets, which spread out in all directions with the speed of the wave.

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New Wavefront: The forward envelope (tangential surface) of these secondary wavelets at any instant gives the new wavefront at that instant.

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Wavefront: A surface of constant phase. For a point source, wavefronts are spherical; for a line source, they are cylindrical; and for a distant source, they are plane wavefronts.

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Reflection of Plane Waves: When a plane wavefront is incident on a reflecting surface, the secondary wavelets from different points reach the surface at different times. Using Huygens construction, it can be proven that the angle of incidence ∠i\angle i is equal to the angle of reflection ∠r\angle r.

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Refraction of Plane Waves: When a plane wave transitions from medium 1 (velocity v1v_1) to medium 2 (velocity v2v_2), the wavefront changes direction. Snell's Law is derived as sin⁡isin⁡r=v1v2\frac{\sin i}{\sin r} = \frac{v_1}{v_2}.

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Effect on Wave Parameters: During refraction, the frequency ν\nu of the light remains constant. However, the speed vv and wavelength λ\lambda change according to the refractive index nn of the medium.

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Refractive Index: The ratio of the speed of light in vacuum (cc) to the speed of light in the medium (vv), expressed as n=cvn = \frac{c}{v}.

📐Formulae

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

sin⁡isin⁡r=v1v2=n2n1=n21\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \frac{n_2}{n_1} = n_{21}

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

v=νλv = \nu \lambda

τ=BCv1=AEv2\tau = \frac{BC}{v_1} = \frac{AE}{v_2}

💡Examples

Problem 1:

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 velocity and wavelength of light in water.

Solution:

Given: λair=589×10−9 m\lambda_{air} = 589 \times 10^{-9}\text{ m}, n=1.33n = 1.33, and c=3×108 m/sc = 3 \times 10^8\text{ m/s}.

  1. Velocity in water: v=cn=3×1081.33≈2.26×108 m/sv = \frac{c}{n} = \frac{3 \times 10^8}{1.33} \approx 2.26 \times 10^8\text{ m/s}
  2. Wavelength in water: λwater=λairn=589 nm1.33≈443 nm\lambda_{water} = \frac{\lambda_{air}}{n} = \frac{589\text{ nm}}{1.33} \approx 443\text{ nm}

Explanation:

The speed of light decreases in a denser medium by a factor of nn. Since frequency is a property of the source and remains constant, the wavelength must also decrease by the same factor nn to satisfy v=νλv = \nu \lambda.

Problem 2:

A plane wave is incident on a glass slab (n=1.5n = 1.5) at an angle of 60∘60^{\circ}. Calculate the angle of refraction using Snell's Law derived from Huygens Principle.

Solution:

Given i=60∘i = 60^{\circ}, n1=1n_1 = 1 (air), n2=1.5n_2 = 1.5. Using Snell's Law: n1sin⁡i=n2sin⁡rn_1 \sin i = n_2 \sin r 1×sin⁡(60∘)=1.5×sin⁡r1 \times \sin(60^{\circ}) = 1.5 \times \sin r 32=1.5sin⁡r\frac{\sqrt{3}}{2} = 1.5 \sin r sin⁡r=32×1.5=1.7323≈0.577\sin r = \frac{\sqrt{3}}{2 \times 1.5} = \frac{1.732}{3} \approx 0.577 r=sin⁡−1(0.577)≈35.2∘r = \sin^{-1}(0.577) \approx 35.2^{\circ}

Explanation:

As light moves from a rarer medium (air) to a denser medium (glass), the speed of the wavefront decreases, causing the refracted wavefront to bend towards the normal, resulting in r<ir < i.