Consider a ray of light incident from air onto a slab of glass (refractive index $n$) of width $d$, at an angle $ heta$. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is

(a) $\dfrac{4\pi d}{\lambda}\left(1 - \dfrac{1}{n^2}

Question

Consider a ray of light incident from air onto a slab of glass (refractive index $n$) of width $d$, at an angle $	heta$. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is

(a) $\dfrac{4\pi d}{\lambda}\left(1 - \dfrac{1}{n^2}\sin^2	hetaight)^{1/2} + \pi$
(b) $\dfrac{4\pi d}{\lambda}\left(1 - \dfrac{1}{n^2}\sin^2	hetaight)^{1/2}$
(c) $\dfrac{4\pi d}{\lambda}\left(1 - \dfrac{1}{n^2}\sin^2	hetaight)^{1/2} + \dfrac{\pi}{2}$
(d) $\dfrac{4\pi d}{\lambda}\left(1 - \dfrac{1}{n^2}\sin^2	hetaight)^{1/2} + 2\pi$

Accepted Answer

> [!TLDR]
> Thin-film reflection: optical path difference $= 2nd\cos r = 2d\sqrt{n^2-\sin^2	heta}$, plus a $\pi$ shift at the top surface, giving $\delta = \frac{4\pi d}{\lambda}\sqrt{n^2-\sin^2	heta} + \pi$ — matching option (a).

**1. Identify the two interfering rays.** Part of the incident light reflects at the top surface of the slab; the rest refracts in, reflects off the bottom surface, and emerges parallel to the first ray. These two reflected rays interfere.

**2. Refraction angle.** By Snell's law at the top face,
$$\sin	heta = n\sin r \quad\Rightarrow\quad \sin r = \frac{\sin	heta}{n}, \qquad \cos r = \sqrt{1 - \frac{\sin^2	heta}{n^2}}.$$

**3. Extra optical path.** The standard thin-film result for the path difference between the top- and bottom-surface reflections is
$$\Delta = 2 n d \cos r.$$
Substituting $\cos r$,
$$\Delta = 2 n d\sqrt{1 - \frac{\sin^2	heta}{n^2}} = 2d\sqrt{n^2 - \sin^2	heta}.$$

**4. Phase from the path.** A path difference $\Delta$ corresponds to phase
$$\delta_{	ext{path}} = \frac{2\pi}{\lambda}\,\Delta = \frac{4\pi d}{\lambda}\sqrt{n^2 - \sin^2	heta}.$$

**5. Reflection phase shift.** The top reflection is at a rarer→denser boundary (air→glass), which flips the wave by $\pi$; the bottom reflection (glass→air) has no such shift. Net extra phase $= \pi$.

**6. Total, and matching to the printed options.**
$$\delta = \frac{4\pi d}{\lambda}\sqrt{n^2 - \sin^2	heta} + \pi = \frac{4\pi n d}{\lambda}\sqrt{1 - \frac{\sin^2	heta}{n^2}} + \pi.$$
All four printed options share the same $\left(1-\frac{1}{n^2}\sin^2	hetaight)^{1/2}$ prefactor (a shared textbook simplification that drops the outer factor of $n$) — they differ only in the additive term. Our derivation gives an additive $+\pi$, which uniquely picks out option (a).

> [!ANSWER]
> **Option (a).** $$\delta = \frac{4\pi d}{\lambda}\left(1-\frac{1}{n^2}\sin^2	hetaight)^{1/2} + \pi.$$

Q.Consider a ray of light incident from air onto a slab of glass (refractive index nnn) of width ddd, at an angle θ\thetaθ. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is

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Q.Consider a ray of light incident from air onto a slab of glass (refractive index $n$ ) of width $d$ , at an angle $\theta$ . The phase difference between the ray reflected by the top surface of the glass and the bottom surface is

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