Deriving Snell's law with Huygens' principle: why the tangent, does frequency change in glass, and why no backward wave?

7 answers & replies

• [+][–]Arjun Nair12thSample· 1mo ago

  Derivation, in two lines. In the time $\tau$ that the wavefront takes to move from $B$ to $C$ in medium 1, $BC = v_1\tau$. In the same time the wavelet from $A$ travels $AE = v_2\tau$ in glass. From the right-angled triangles with common hypotenuse $AC$:

  $\sin i = \frac{BC}{AC} = \frac{v_1\tau}{AC},\qquad \sin r = \frac{AE}{AC} = \frac{v_2\tau}{AC}$

  Dividing, $\dfrac{\sin i}{\sin r} = \dfrac{v_1}{v_2} = \dfrac{n_2}{n_1}$. In glass $v_2 < v_1$, so $\sin r < \sin i$ and $r < i$, which is bending towards the normal.

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• [+][–]Sara Fernandes11thSample· 1mo ago

  Small number to feel it: refractive index $n = c/v$. For glass with $n = 1.5$, $v = 3\times10^8/1.5 = 2\times10^8$ m/s. So $v_2/v_1 = 2/3$ and Snell's law gives $\sin i/\sin r = 1.5$. For your Q3, light crossing from air into glass slows down by that factor, hence the front turns towards the normal. I'm class 11 so I might be missing something, but this is how my teacher showed it.

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• [+][–]Zoya Khan12thSample· 1mo ago

  On your Q2, the answer is that in refraction the frequency stays the same, because it's fixed by the source. Think of the boundary: the number of wave crests arriving per second at the surface has to equal the number leaving into the glass, otherwise crests would pile up or vanish at the boundary. So $f$ is constant and from $v = f\lambda$, the wavelength gets shorter in glass: $\lambda_2 = \lambda_1 \times v_2/v_1$, which for light coming from air (about vacuum) is $\lambda_0/n$.

  For (1): the wavelets reinforce each other only along their common tangent, where they all arrive in phase. Everywhere else they cancel, so that surface is the wavefront.

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  • [+][–]Lakshmi TripathiDropperSample· 1mo agoAuthor

    the muddy marching band picture and the "frequency fixed by source" line were what I needed. thank you both, really. I will try asking in class next time 🙂

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  • [+][–]Sara Fernandes11thSample· 1mo ago

    ohh crests can't pile up at the boundary, so f is fixed. thanks Zoya, that's neat

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• [+][–]Tushar PawarDropperSample· 1mo ago

  Intuition for (3): picture a marching band hitting a muddy patch at an angle. The side that enters the mud first slows down, the other side is still fast, so the whole line pivots towards the normal. A wavefront is that line, with the part in glass lagging. Because the end that has entered glass travels less distance in the same time, the front tilts.

  Shape of wavefronts for NEET: a point source gives spherical wavefronts, a very distant source gives plane wavefronts, and a linear (slit) source gives cylindrical wavefronts.

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• [+][–]Mehak AroraDropperSample· 1mo ago

  For (4): in the simple Huygens construction we only draw the forward envelope, which can seem like cheating. NCERT mentions that the amplitude of the secondary wavelet is maximum in the forward direction and zero in the backward direction; that was shown later by Fresnel. For exams, just say we consider the forward envelope.

  Huygens' principle also explains reflection (gives $i = r$) and that light rays are perpendicular to wavefronts, good to know for the 2–3 mark questions.

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