# Effect of incident angle on wavelength of transmitted wave for normal polarisation?

In my electrodynamcis assignment I'm being asked to derive the wavelength of a normally polarised wave transmitted through a glass/air interface as a function of $$n_1$$ (the refractive index of the first medium) using the concept of phase continuity. I've tried to derive it and keep getting $$\lambda_t = \lambda_i \, n1/n2$$. ($$n2=1$$ because medium 2 is air)

I've been asked to do this for the critical angle and for an angle of incidence equal to $$\pi/3$$, but I don't see how there can be a difference if Snell's law causes the incident angle to cancel out? We've been doing derivations from first principles but I don't see how to get to an equation for $$\lambda_t$$ without using Snell's law and the phase continuity condition that the maxima of the incident and transmitted waves must line up. Maybe there's a more complex equation for phase continuity?

Any help would be appreciated, thank you!

The ratio if incident to transmitted wavelengths and Snell's law both arise from the requirement of phase continuity, since the arguments of the incident, reflected and transmitted waves must all be the same. i.e. $$\vec{k_i}\cdot \vec{r} - \omega_i t = \vec{k_r}\cdot \vec{r} - \omega_r t = \vec{k_t}\cdot \vec{r} - \omega_t t\$$ for any value of $$\vec{r}$$ along the interface and for any value of $$t$$, where the $$i$$, $$r$$ and $$t$$ subscripts refer to incident, reflected and transmitted electric field.
From the temporal part of the argument (i.e. for fixed $$\vec{r}$$), this leads to a continuity of frequency and hence a discontinuity in wavelength $$\frac{c}{n_i \lambda_i} = \frac{c}{n_t \lambda_t}\ ,$$ where the incidence/transmission angles do not feature; and to Snell's law from the continuity of the spatial part of the argument (i.e. at fixed $$t$$) $$\frac{\sin\theta_i}{\lambda_i} = \frac{\sin\theta_t}{\lambda_t}\ ,$$ where $$\theta_i$$ is the arbitrary angle of incidence.
• @Veronica I have shown that $\lambda_t = (n_t/n_i)\lambda_i$ without using Snell's law, which in your question you said you could not do. The phase continuity condition implies that the arguments of the sinusoidally varying (in space and time) electric fields parallel to the interface must be the same for any position along the interface at any time. To accomplish this means that $\omega_i t = \omega_t t$ and hence is independent of the incidence angle. Commented Apr 10 at 17:02