Refraction at Plane Surfaces

Lateral shift is the perpendicular distance between the direction of the incident ray and the direction of the emergent ray when a ray passes obliquely through a parallel-sided glass slab.

Setup. A ray $AO$ is incident at $O$ on the upper face of a slab of thickness $t$ and refractive index $\mu$, with angle of incidence $i$. It is refracted along $OB$ with angle of refraction $r$, reaches the lower face at $B$ and emerges along $BC$, parallel to $AO$ because the faces are parallel.

Construction. Produce $AO$ straight on to a point $N$ so that $BN \perp ON$. Then $BN = d$ is the lateral shift. Let $M$ be the foot of the perpendicular from $B$ to the upper face's normal at $O$, so $OM = t$ along the normal.

Geometry.

• In right triangle $OMB$, $\angle MOB = r$, so $OB = \frac{t}{\cos r}$
• The undeviated direction $ON$ makes angle $i$ with the normal, and $OB$ makes angle $r$ with it on the same side, so $\angle NOB = i - r$.
• In right triangle $ONB$, $d = BN = OB\sin(i-r)$

Result.

$\boxed{d = \frac{t\sin(i-r)}{\cos r}}$

Expanding $\sin(i-r)=\sin i\cos r-\cos i\sin r$,

$d = t\sin i\left(1-\frac{\cos i}{\mu\cos r}\right)=t\sin i\left(1-\frac{\cos i}{\sqrt{\mu^2-\sin^2 i}}\right)$

using $\sin r = \dfrac{\sin i}{\mu}$ so that $\mu\cos r=\sqrt{\mu^2-\sin^2 i}$.

For small angles, $d \approx t\, i\left(1-\dfrac1\mu\right)$. The lateral shift increases with the thickness $t$, the angle of incidence $i$ and the refractive index $\mu$; it is zero for normal incidence.

(d) glass to water

Total internal reflection needs light to travel from a denser to a rarer medium at an angle of incidence greater than the critical angle. Glass ($n\approx1.5$) is optically denser than water ($n\approx1.33$), so only the glass-to-water case qualifies.

(a) $2\times10^{8}$ m/s

$v=\frac{c}{n}=\frac{3\times10^{8}}{1.5}=2\times10^{8}\ \text{m/s}$

Lateral shift: the perpendicular distance between the path of the incident ray (extended) and the emergent ray when light passes through a parallel-sided glass slab. For slab thickness $t$, incidence angle $i$ and refraction angle $r$,

$d=\frac{t\sin(i-r)}{\cos r}$

Variation with angle of incidence: At $i=0^\circ$ the ray goes straight through and $d=0$. As $i$ increases, $d$ increases steadily, because $i-r$ grows. For grazing incidence ($i\to90^\circ$) the shift reaches its maximum value, $d\to t$. The graph of $d$ against $i$ is therefore a curve rising from the origin, gently at first and then more steeply, up to $d=t$ at $90^\circ$.

No dispersion by a slab: The two faces of the slab are parallel. Different colours are refracted by different amounts at the first face, so they separate inside the glass. At the second face each colour is refracted back by the same angle in the opposite sense, so every colour emerges parallel to the original incident direction. The emergent rays are parallel and so overlap almost completely (they are separated only by a very small lateral distance), and the eye sees white light. Without an angle between the faces, as in a prism, there is no net angular spread of colours.

By Snell's law, $\mu=\dfrac{\sin i}{\sin r}$:

$\sin r=\frac{\sin60^\circ}{1.5}=\frac{0.866}{1.5}=0.577$ $r=\sin^{-1}(0.577)\approx35.3^\circ$

The angle of refraction in the glass is about $35^\circ$.

(a) The refractive index of a material depends on (i) the nature of the material (medium), and (ii) the wavelength (colour) of the light; it also varies slightly with temperature.

(b) Refractive index of ice:

$\mu=\frac{\sin i}{\sin r}=\frac{\sin45^\circ}{\sin30^\circ}=\frac{0.7071}{0.5}=1.414$

Critical angle:

$\sin C=\frac{1}{\mu}=\frac{1}{1.414}=0.7071\ \Rightarrow\ C=45^\circ$

(c) For a ray in ice incident on the ice-air boundary at $45^\circ$, the angle of incidence equals the critical angle. The refracted ray therefore grazes along the boundary, with an angle of refraction of $90^\circ$ (it is just on the verge of total internal reflection).

(b) (\sqrt{2})

For total internal reflection, the refractive index of the denser medium with respect to the rarer medium is $n=\frac{1}{\sin C}=\frac{1}{\sin 45^\circ}=\frac{1}{1/\sqrt2}=\sqrt2$

Total internal reflection is the phenomenon in which a ray of light travelling from a denser medium to a rarer medium, striking the interface at an angle greater than the critical angle, is completely reflected back into the denser medium instead of being refracted.

Necessary conditions:

1. Light must travel from a denser medium to a rarer medium (e.g. glass to air, water to air). If light goes from rarer to denser, total internal reflection cannot occur.
2. The angle of incidence in the denser medium must be greater than the critical angle for the pair of media. The critical angle $C$ is given by $\sin C = \frac{1}{n}$ where $n$ is the refractive index of the denser medium with respect to the rarer medium. At $i = C$, the refracted ray grazes along the interface ($r = 90^\circ$); for $i > C$, no refracted ray exists and all the light is reflected.

Use in optical fibres:

An optical fibre is a thin, flexible thread of high-quality glass or plastic consisting of a central core of higher refractive index surrounded by a cladding of lower refractive index. Light entering one end of the fibre at a small angle strikes the core–cladding boundary at an angle greater than the critical angle. Since the conditions for total internal reflection are satisfied, the light is repeatedly and totally reflected along the length of the fibre with practically no loss of intensity. In this way, light signals (carrying telephone, internet and TV data) travel long distances along the fibre, even around bends, and emerge at the other end. Optical fibres are thus used in communication, endoscopy and other optical instruments.

Laws of refraction

1. The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
2. Snell's law: for a given pair of media and a given colour of light, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant:

$\frac{\sin i}{\sin r} = {}_1\mu_2 = \text{constant}$

Refractive index

The refractive index of a medium is the ratio of the speed of light in vacuum to its speed in the medium:

$\mu = \frac{c}{v}$

Equivalently, for light going from air into the medium, $\mu = \dfrac{\sin i}{\sin r}$. It has no unit. For example, $\mu_{\text{water}}\approx1.33$ and $\mu_{\text{glass}}\approx1.5$.

Why a pencil looks bent in water

Light from the immersed part of the pencil passes from water (denser) to air (rarer) and bends away from the normal. The eye traces the rays back in straight lines, so the immersed part appears at a smaller depth than its real position. The part above water is seen at its true position, so the pencil appears bent (and shortened) at the water surface.

Apparent depth $=\dfrac{1}{\mu}\times$ real depth

Let $O$ be a point object at real depth $h$ below the surface of water (refractive index $\mu$). Let $B$ be the point of the surface vertically above $O$. A ray $OA$ meets the surface at $A$, very close to $B$, and is refracted into air away from the normal. Seen from above, the object appears at $I$, where the refracted ray traced backwards meets $OB$. Let $BI=h'$ be the apparent depth.

Angle of incidence in water: $\angle AOB = i$. Angle of refraction in air: $\angle AIB = r$.

$\tan i=\frac{AB}{h},\qquad \tan r=\frac{AB}{h'}$

Light goes from water to air, so $\mu\sin i=\sin r$, i.e.

$\mu=\frac{\sin r}{\sin i}$

For small angles $\sin\approx\tan$, so

$\mu\approx\frac{\tan r}{\tan i}=\frac{AB/h'}{AB/h}=\frac{h}{h'}$

$\boxed{h'=\frac{h}{\mu}}$

Hence the apparent depth is $\dfrac1\mu$ times the real depth. Since $\mu>1$, the object appears raised.