Refraction at Plane Surfaces & Total Internal Reflection

PHYRefraction at Plane Surfaces & TIR

When light crosses a plane boundary between two media it bends according to Snell's law, with the bending governed by the refractive index $\mu = c/v$ — the ratio of the speed of light in vacuum to its speed in the medium. This article covers the everyday consequences of that bending: how a slab makes objects look shallower (apparent depth), how it shifts an emergent ray sideways (lateral displacement), and how, beyond a critical angle, light is trapped entirely inside the denser medium (total internal reflection, TIR).

These are heavily examined because one idea — Snell's law — ties together real devices like glass slabs, prisms and optical fibres, and the same $\mu$ also fixes the speed $v$ in the medium and (via the single-surface relation) image formation at a curved interface.

Snell's law and refractive index

{}_1\mu_2 = \dfrac{\sin i}{\sin r} = \dfrac{v_1}{v_2} = \dfrac{\lambda_1}{\lambda_2} = \dfrac{\mu_2}{\mu_1}

i

is the angle of incidence and

r

the angle of refraction, both measured from the normal;

v

is wave speed,

\lambda

wavelength;

\mu = c/v

with

c

the speed of light in vacuum. Frequency is unchanged on refraction.

Real and apparent depth

\mu = \dfrac{\text{real depth}}{\text{apparent depth}}

Valid for near-normal viewing from the rarer medium (air). For an object at real depth

d

in a medium of index

\mu

, the apparent depth is

d/\mu

and the object appears raised by

d\left(1-\dfrac{1}{\mu}\right)

Lateral displacement through a parallel slab

d = \dfrac{t\,\sin(i - r)}{\cos r}

t

is the slab thickness,

i

the angle of incidence on the first face,

r

the angle of refraction inside (

\sin i = \mu \sin r

). The emergent ray is parallel to the incident ray but shifted sideways by

d

Critical angle and total internal reflection

\sin C = \dfrac{1}{\mu} = \dfrac{\mu_2}{\mu_1}

C

is the critical angle for a denser medium of index

\mu_1

in contact with a rarer medium

\mu_2

(here

\mu = \mu_1/\mu_2

). TIR occurs only when light goes from denser to rarer and the angle of incidence exceeds

C

Acceptance angle of an optical fibre

\sin i_{max} = \sqrt{\mu_{core}^{2} - \mu_{clad}^{2}}

i_{max}

is the maximum half-angle (acceptance angle) in air for which a ray entering the core face still undergoes TIR at the core–cladding boundary;

\mu_{core}

and

\mu_{clad}

are the core and cladding indices, with air (

\mu = 1

) outside. The quantity

\sin i_{max}

is the numerical aperture.

Refraction at a single spherical surface

\dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R}

Light travels from medium

\mu_1

(object side) into

\mu_2

;

u

and

v

are object and image distances and

R

the radius of curvature, all measured from the pole with the Cartesian sign convention (distances against incident light negative).

Medium (against air)	Refractive index $\mu$	Critical angle $C$
Water	$1.33$	$48.8^{\circ}$
Crown glass	$1.50$	$41.8^{\circ}$
Dense flint glass	$1.65$	$37.3^{\circ}$
Diamond	$2.42$	$24.4^{\circ}$

Critical angles from

\sin C = 1/\mu

. The denser the medium, the smaller the critical angle — which is why a diamond, with

C \approx 24^{\circ}

, traps and returns so much of the light that enters it.

Attacking any total-internal-reflection question

1Check the direction: TIR is possible only when light goes from the denser medium into the rarer one. Air into water can never give TIR, whatever the angle.
2Find the critical angle from $\sin C = \dfrac{\mu_{\text{rarer}}}{\mu_{\text{denser}}}$ , which is $\dfrac{1}{\mu}$ when the rarer medium is air.
3Compare the angle of incidence at the interface with $C$ : greater than $C$ gives total reflection, exactly $C$ gives grazing emergence, less than $C$ gives ordinary refraction.
4For a source under a liquid, the light that escapes forms a circle of radius $r = h\tan C$ ; everything outside that circle is totally reflected.

Refractive index sets the speed: $v = c/\mu$ , so a higher $\mu$ means slower light. Frequency stays constant across a boundary, while wavelength changes as $\lambda_{medium} = \lambda_{vacuum}/\mu$ .
Snell's law uses sines of the angles measured from the normal, never from the surface. Light bends toward the normal entering a denser medium and away from the normal entering a rarer one.
The real-depth/apparent-depth rule $\mu =$ real/apparent assumes viewing nearly along the normal, so that the small-angle approximation $\sin\theta \approx \tan\theta$ holds. For several stacked slabs the apparent depths add: total apparent depth $= \dfrac{d_1}{\mu_1} + \dfrac{d_2}{\mu_2} + \cdots$ , where $d_i$ is the real thickness of each layer of index $\mu_i$ .
In a parallel-sided slab the ray emerges parallel to its original direction (no net deviation) but laterally shifted; the shift $d$ grows with thickness $t$ and with the angle of incidence, vanishing at normal incidence ( $i = 0$ ).
TIR has two strict conditions: light must travel from the optically denser to the rarer medium, AND the angle of incidence must exceed the critical angle $C$ . A larger $\mu$ gives a smaller $C$ , so denser media trap light more easily.
A $45^{\circ}$ – $90^{\circ}$ – $45^{\circ}$ glass prism reflects light by TIR when $\mu > \sqrt{2} \approx 1.414$ , since then $C < 45^{\circ}$ ; this is how totally-reflecting prisms replace mirrors in periscopes and binoculars.
In a fibre, light entering within the acceptance cone hits the core–cladding wall beyond its critical angle and is guided by repeated TIR; the acceptance angle is found by combining Snell's law at the entry face with the TIR condition inside.
For refraction at a single spherical surface use $\dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R}$ with the same Cartesian sign convention as mirrors/lenses — fix the sign of $R$ from whether the centre of curvature lies on the outgoing-light side ( $R$ positive) or not.

Where the marks go

Sign-convention slips at a spherical surface: students plug raw positive numbers into $\dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2-\mu_1}{R}$ . With light incident from the left, $u$ is negative (object against incident light) and $R$ is positive only if the centre of curvature lies on the outgoing-light side. Decide $\mu_1, \mu_2$ by direction of travel, not by which medium is 'glass'.
Mixing up which way $\mu$ goes in apparent depth: apparent depth $=$ real depth $/\mu$ (object looks shallower), so $\mu$ DIVIDES the real depth. Writing apparent $= \mu \times$ real makes a denser medium look deeper, which is wrong.
Forgetting the direction requirement for TIR: applying $\sin C = 1/\mu$ when light is going from rarer to denser. TIR is impossible in that direction — there is always a refracted ray. Also, TIR proper needs $i > C$ : exactly at $i = C$ the refracted ray grazes along the surface at $90^{\circ}$ , so $i = C$ is the threshold, not the TIR regime itself.
Confusing $\sin$ with $\tan$ and using the wrong angle in lateral shift: the displacement is $d = \dfrac{t\sin(i-r)}{\cos r}$ , with $r$ from Snell's law $\sin i = \mu\sin r$ — not the geometric face angle. Students also forget that lateral shift causes zero net deviation, unlike a prism.

How the board asks it

Numericalcritical angle and total internal reflection
The refractive index of glass is $1.5$ . Calculate the critical angle for the glass-air interface, and state whether a ray striking the surface at $45^\circ$ from inside the glass undergoes total internal reflection.
Derive / provelateral displacement through a parallel slab
A ray of light is incident at an angle $i$ on a parallel-sided glass slab of thickness $t$ and refractive index $\mu$ . Derive an expression for the lateral displacement of the emergent ray and show that it is $d = \dfrac{t\sin(i-r)}{\cos r}$ , where $r$ is the angle of refraction.
Numericalreal and apparent depth for stacked layers
A tank holds a $24\,\text{cm}$ layer of water ( $\mu = 1.33$ ) over a $9\,\text{cm}$ layer of oil ( $\mu = 1.5$ ). Calculate the apparent depth of the bottom of the tank as seen by an observer looking vertically downwards.
Give reasonstir direction condition1 mk
Give reasons why total internal reflection cannot occur when light travels from air into water, however large the angle of incidence may be.
Diagram / graphtotally-reflecting prism turning a ray through 90 degrees
Draw a labelled ray diagram to show how a right-angled isosceles glass prism turns a ray of light through $90^\circ$ by total internal reflection, marking the critical angle and the path of the ray.
Numericalrefraction at a single spherical surface
A point object in air is placed $30\,\text{cm}$ in front of the convex spherical surface of a glass medium ( $\mu = 1.5$ ) of radius of curvature $20\,\text{cm}$ . Using $\dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R}$ , find the position of the image.
Define / statenormal incidence, and matched refractive indices1 mkAsked 2026
Give any one example where a ray of light travelling from one optical medium into another travels undeviated.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.