A plane wavefront travelling in a rarer medium of speed $v_1$ is incident on the plane boundary of a denser medium of speed $v_2$ (with $v_2

Question

A plane wavefront travelling in a rarer medium of speed $v_1$ is incident on the plane boundary of a denser medium of speed $v_2$ (with $v_2<v_1$) at an angle of incidence $i$. Using Huygens' construction, derive the relation $\sin i/\sin r=v_1/v_2$ connecting the angle of refraction $r$ to the two speeds, and explain why the refracted wavefront bends towards the normal.

Accepted Answer

Let the plane wavefront $AB$ in medium 1 (speed $v_1$ ) be incident on the interface $MN$ at $A$ , at angle of incidence $i$ . In time $\tau$ , point $B$ reaches the interface at $C$ , so $BC=v_1\tau$ . In the same time, the secondary wavelet sent out from $A$ into medium 2 (speed $v_2$ ) has grown to radius $AE=v_2\tau$ . The refracted wavefront is the tangent $CE$ , making angle of refraction $r$ with the normal.

In right triangle $ABC$ : $\sin i=\dfrac{BC}{AC}=\dfrac{v_1\tau}{AC}$ .

In right triangle $AEC$ : $\sin r=\dfrac{AE}{AC}=\dfrac{v_2\tau}{AC}$ .

Dividing the two relations cancels the common factor $\tau/AC$ :

\frac{\sin i}{\sin r}=\frac{v_1}{v_2}

Since medium 2 is denser, $v_2<v_1$ , so $\sin i/\sin r>1$ , i.e. $\sin i>\sin r$ , i.e. $i>r$ . The refracted wavefront therefore bends towards the normal on entering the denser medium -- physically, because the part of the wavefront that has already entered the slower medium is 'held back' relative to the part still in the faster medium, tilting the wavefront (and hence the ray, perpendicular to it) closer to the normal.

✓Final answer

\sin i/\sin r=v_1/v_2

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2023Set ANNUAL5 marks
$Q.(a) Using Huygens' construction, draw a figure showing the propagation of plane wave refracting at a plane surface separating two media. Hence, verify Snell's law of refraction. (1+2)$
(b) An unpolarized beam of light is incident on a group of three polarising sheets, which are arranged in such a way that the characteristic direction of each polarising sheet makes an angle of 30° with that of the preceding sheet. What fraction of light is transmitted?
(2) OR
(a) In Young's double slit experiment, deduce the conditions for
(i) constructive and
(ii) destructive interference at a point on the screen. (2+1+1)
(b) Draw a graph showing variation of the resultant intensity in the interference pattern against position on the screen. (1)
›Reveal solutionSolution
$Figure — The answered alternative uses Huygens' construction to refract a plane wavefront at a plane interface and veri$
Figure — The answered alternative uses Huygens' construction to refract a plane wavefront at a plane interface and veri

Huygens' principle constructs the refracted wavefront from secondary wavelets, whose geometry directly reproduces Snell's law; passing unpolarised light through 3 polarisers each rotated 30° from the previous one transmits 9/32 of the original intensity.

Huygens' construction and Snell's law (described): Consider a plane wavefront AB incident obliquely on a plane interface XY separating medium 1 (speed $v_1$ ) from medium 2 (speed $v_2$ ), making angle of incidence $i$ with the normal. According to Huygens' principle, every point on the wavefront acts as a source of secondary spherical wavelets that spread out with the speed appropriate to the medium they are in. At the instant point A on the wavefront touches the interface, point B (also on the incident wavefront) still has to travel a distance $BC$ (at speed $v_1$ ) to reach the interface, taking time $\tau = BC/v_1$ . During this same time $\tau$ , the secondary wavelet from A (already at the interface, now inside medium 2) spreads out to a sphere of radius $AD = v_2\tau$ . The new refracted wavefront is the common tangent CD drawn from C to this expanding wavelet. From the right triangles $ABC$ (angle $i$ at A) and $ACD$ (angle $r$ at C, the angle of refraction), with $AC$ as the common hypotenuse: $\sin i = \dfrac{BC}{AC} = \dfrac{v_1\tau}{AC}, \qquad \sin r = \dfrac{AD}{AC} = \dfrac{v_2\tau}{AC}$ Dividing: $\dfrac{\sin i}{\sin r} = \dfrac{v_1}{v_2} = n_{21}$ This is Snell's law of refraction, where $n_{21}$ is the refractive index of medium 2 relative to medium 1 — derived purely from the geometry of Huygens' wavelets.

Fraction of light transmitted through three polarisers: When unpolarised light of intensity $I_0$ passes through the first polariser, it becomes polarised and its intensity is halved (regardless of the polariser's orientation), by the fundamental property of a polariser acting on unpolarised light: $I_1 = \dfrac{I_0}{2}$ The now-polarised light then passes through the second polariser, whose transmission axis is at $30°$ to the first. By Malus's law, $I = I_0\cos^2\theta$ : $I_2 = I_1\cos^2(30°) = I_1\times\dfrac34$ Similarly, passing through the third polariser (at another $30°$ relative to the second): $I_3 = I_2\cos^2(30°) = I_2\times\dfrac34$ Combining: $I_3 = \dfrac{I_0}{2}\times\dfrac34\times\dfrac34 = \dfrac{9I_0}{32}$ So the fraction of the original intensity transmitted is: $\dfrac{I_3}{I_0} = \dfrac{9}{32} \approx 0.281\ (28.1\%)$

✓Final answer
$Huygens' construction gives sin i/sin r = v₁/v₂ = n₂₁, which is Snell's law. Fraction transmitted = 9/32 \approx 28.1%.$

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