Total internal reflection | ICSE Class 10 Physics Notes

This note covers total internal reflection, optical density, critical angle, the relation between critical angle and refractive index, ray paths through triangular glass prisms, deviations of 90° and 180°, comparison with plane mirrors, and optical fibres.

What happens when light travels from a denser medium to a rarer medium?

How are the media and angles described?

A medium is a material through which light travels. A ray represents the direction in which light travels. The boundary between two media is their interface. At this boundary, light may return into its original medium by reflection or enter the other medium by refraction.

The normal is a line perpendicular to the interface at the point where the ray strikes it. The arriving ray is the incident ray. Its angle with the normal is the angle of incidence, written as i. Angles here are measured in degrees, represented by °.

The ray entering the second medium is the refracted ray. Its angle with the normal is the angle of refraction, written as r. Both angles are measured from the normal, not from the surface.

An optically denser medium has a higher refractive index than the medium with which it is compared. Refractive index describes the relative slowing of light in a medium. An optically rarer medium has the lower refractive index.

Which way does the refracted ray bend?

For oblique incidence, meaning incidence at an angle between 0° and 90° to the normal, light travelling from water or glass into air bends away from the normal. Therefore, r is greater than i. Some light is also reflected back into the denser medium.

As i increases, r also increases. The refracted ray comes progressively closer to the boundary. This change leads to the limiting situation called the critical angle, followed by total internal reflection.

Note: Optical density is not mass density. Mass density means mass per unit volume; optical density concerns the speed of light. Do not decide which material is optically denser by asking which material is heavier.

What is total internal reflection and when does it occur?

Definition: Total internal reflection is the complete reflection of light back into the optically denser medium when it reaches a boundary with an optically rarer medium at an incidence angle greater than the critical angle.

The critical angle, represented by C, is the incidence angle in the denser medium for which the refracted ray in the rarer medium makes 90° with the normal. At this angle, the refracted ray travels along the interface.

What are the two essential conditions?

1. Light must approach the boundary from the optically denser medium towards the optically rarer medium.
2. The angle of incidence at that boundary must be greater than the critical angle for the pair of media.

Both conditions are necessary. A large incidence angle alone does not establish total internal reflection. A ray travelling from air into glass approaches the boundary from the rarer side and therefore does not satisfy the first condition.

Likewise, a ray travelling from glass towards air does not necessarily undergo total internal reflection. Its incidence angle must also be checked. Below C, some light crosses the boundary; at C, the refracted ray grazes the boundary; above C, total internal reflection occurs.

Why does the word “internal” matter?

The reflected ray stays inside the medium from which it approached the boundary. For example, a ray approaching a water-air boundary from within water returns into water. “Internal” identifies where the ray travels, while “total” distinguishes the process from partial reflection.

The law of reflection still applies: the angle made by the reflected ray with the normal equals the angle of incidence. The incident ray, reflected ray and normal lie in the same plane. Total internal reflection changes the amount reflected, not this geometrical rule.

How does the critical angle separate the three possible ray paths?

Keep the pair of media fixed and increase the angle of incidence in the denser medium. There are three distinct cases. The boundary between partial transmission and total internal reflection must be identified precisely.

How should the boundary diagram be read?

Water occupies the lower region and air the upper region. Rays from point A in water meet the boundary at different points. The figure shows refracted rays in air, a ray along the boundary, partially reflected rays in water and a totally reflected ray.

See Fig. 9.11 in your NCERT textbook

The figure compares different rays from the same point. At each point of incidence, use the normal drawn there. The water-air boundary is horizontal, so its normals are vertical. An angle drawn to the horizontal boundary is not the incidence angle.

At the critical angle, r = 90°. This does not mean that i is 90°. The critical angle is measured inside the denser medium and is smaller than 90° for the situations considered here.

Above the critical angle, do not extend a supposed refracted ray into air. Instead, draw the reflected ray inside water, on the other side of the normal, making an angle equal to i. An arrow should show that the reflected ray travels away from the boundary.

How are critical angle and refractive index related?

Let μ, pronounced “mu”, represent the refractive index of the denser medium relative to the rarer medium. For a material surrounded by air, use its refractive index relative to air. The symbol sin denotes the sine trigonometric function of an angle.

For refraction from the denser to the rarer medium, Snell’s law can be written as μ sin i = sin r. This relation applies for a fixed pair of media and a fixed colour of light, while a refracted ray exists.

Derivation: Relation between μ and C

1. At the critical angle, substitute i = C into the refraction relation.
2. The refracted ray grazes the boundary, so r = 90°.
3. Since sin 90° = 1, the relation becomes μ sin C = 1.

μ = 1/sin C

The equivalent form is sin C = 1/μ. Refractive index is a ratio and has no unit. Here angles are expressed in degrees, shown by the symbol °. When evaluating a sine or its inverse, use degree mode if the angle is required in degrees.

What does the relation tell us?

For materials compared with the same rarer medium, a higher refractive index corresponds to a smaller critical angle. A smaller critical angle allows total internal reflection at a smaller incidence angle, provided the ray is travelling towards the rarer medium.

The relation depends on the pair of media. A critical angle quoted for glass and air cannot simply be reused for glass and water. Changing the surrounding medium changes the relative refractive index and therefore the limiting angle.

At incidence angles greater than C, Snell’s law would require a sine of the refraction angle greater than one. No real refraction angle can meet that requirement. This is the mathematical reason that the transmitted ray disappears in the geometrical description.

How large is the surface through which an underwater bulb can send light?

Worked example 3. A small bulb lies at the bottom of a tank filled with water to a depth of 80 cm. Treat the bulb as a point source and take the refractive index of water as 1.33. Find the area of the water surface through which its light can emerge.

Answer: Rays reaching the surface below the critical angle can emerge. The limiting rays define a circle centred vertically above the bulb. Let $h$ be the depth, $R$ the circle's radius, $S$ its area and $C$ the critical angle.

Formula: $\sin C = 1/\mu$, $R = h\tan C$ and $S = \pi R^2$. The right triangle between the bulb, the circle's centre and its edge gives the radius relation.

Substitute: $C = \sin^{-1}(1/1.33) \approx 48.7535^\circ$. Then $R = 80\tan(48.7535^\circ) \approx 91.23\,\mathrm{cm}$.

The limiting radius is approximately 91.23 cm. Using the unrounded radius, $S = \pi[80\tan C]^2 \approx 26149.3\,\mathrm{cm}^2$, so the area is approximately 2.61 m². At the circular boundary the refracted ray grazes the surface; rays reaching beyond it undergo total internal reflection.

What do the tabulated critical angles tell us?

The following values compare transparent materials with air. The critical-angle entries are in degrees. Keep each refractive index paired with its listed critical angle; the table gives values at the precision shown.

Within this table, refractive index increases from water to diamond while critical angle decreases. Do not substitute one row’s refractive index into a calculation intended for another material. “Glass” alone does not specify a single exact refractive index.

How can the values decide whether reflection is total?

Worked example 1. A ray inside crown glass approaches air at i = 45°. Its critical angle is 41.14°. Decide whether total internal reflection occurs.

Answer: Crown glass is optically denser than air. Since 45° is greater than 41.14°, both conditions are satisfied. The ray undergoes total internal reflection and its angle of reflection is 45°.

Worked example 2. A 45°, 45°, 90° dense-flint-glass prism in air is used in the reversing arrangement, with 45° incidence at each of its two reflecting faces. Its critical angle is 37.31°. Check both reflections.

Answer: At each face, the ray approaches air from denser glass and 45° exceeds 37.31°. Both reflections are total internal reflections. The two-reflection arrangement reverses the ray direction, producing 180° deviation.

Both crown glass and dense flint glass therefore support the standard right-angle prism arrangements in air. The critical-angle check explains why these materials work: their values fall below the 45° incidence at the internal reflecting faces.

For crown glass, an internal incidence of 30° is below the listed critical angle, whereas 60° is above it. These comparisons are useful when tracing rays through triangular prisms. The angles arise from the prism geometry, rather than from the label “prism” alone.

How does a right-angle prism turn light through 90°?

A triangular prism is a transparent optical block with a triangular cross-section. Its flat boundary surfaces are called faces. A right-angle isosceles prism has cross-section angles 45°, 45° and 90°. Its hypotenuse face corresponds to the side opposite the right angle.

The angle of deviation, represented by δ, is the change between the original direction of the incident ray and the final direction of the emergent ray. The emergent ray is the ray leaving the prism. Deviation is different from incidence at an individual face.

What happens at the three faces?

1. Direct the incident ray normally onto one of the two mutually perpendicular faces. Normal incidence means that the ray follows the normal, so its incidence angle is 0°.
2. The ray enters the glass without bending and reaches the hypotenuse face at an internal incidence angle of 45°.
3. If the glass-air critical angle is less than 45°, the ray undergoes total internal reflection at the hypotenuse face.
4. The reflected ray reaches the other perpendicular face normally and leaves without bending. Its final direction is at right angles to its initial direction.

Thus, δ = 90°. Only the middle encounter is total internal reflection. The entry and exit encounters are transmission through the boundary at normal incidence.

Two horizontal rays enter a 45°, 45°, 90° prism from the left. They reflect at its sloping face and emerge vertically downwards. The triangular section and the right-angle turn are shown.

See Fig. 9.13a in your NCERT textbook

Crown glass and dense flint glass satisfy the required critical-angle condition in air. Their listed critical angles are both below 45°. However, the shape alone is not enough: the material and the medium outside the reflecting face must also permit total internal reflection.

When drawing the path, mark the normal at the reflecting face before marking 45°. The angle between the ray and the face happens to be 45° here too, but incidence is still defined using the normal.

How does the same prism turn light through 180°?

A 45°, 45°, 90° prism can also reverse the direction of a ray. For this arrangement, light enters normally through the hypotenuse face instead of through one of the perpendicular faces. The internal path then includes two reflections.

How is the reversing path traced?

1. Draw the incident ray perpendicular to the hypotenuse face, entering at a point that lets it strike one of the shorter faces rather than a vertex.
2. Continue it straight inside the glass. At the first shorter face, its incidence angle is 45°.
3. For a glass-air critical angle below 45°, draw the first totally reflected ray towards the other shorter face.
4. At the second shorter face, the incidence angle is again 45°. Draw a second total internal reflection.
5. The ray returns to the hypotenuse face normally and emerges travelling opposite to its original direction.

The result is δ = 180°. The incident and emergent rays are parallel but travel in opposite directions. They need not lie on the same straight line: direction reversal does not mean exact retracing of the incoming path.

Two horizontal rays enter the vertical face of a right-angle prism. They reflect at its two sloping faces and return towards the left. The diagram marks the 45° and 90° prism angles and shows the outgoing rays below the incoming rays.

See Fig. 9.13b in your NCERT textbook

For both arrangements, the internal incidence at each reflecting face is 45°. The critical angle must therefore be less than 45°, not merely equal to it. At equality, the refracted ray would graze the face instead of giving the required total internal reflection.

How are rays traced through the other triangular prisms?

Prisms with angles 60°, 60°, 60° and 30°, 60°, 90° require the same boundary-by-boundary reasoning. State the entry face and the direction of entry before deciding which internal angles arise. A triangle’s angles alone do not specify one unique incident ray.

Derivation: Relation between the internal angles of a prism

Let $A$ be the angle between the two prism faces. Let $r_1$ be the refracted angle at entry and $r_2$ the internal incidence angle at the next face. Both are measured from their respective normals. Let $\theta$ be the angle between the normals inside the geometrical construction.

1. The quadrilateral formed by the prism vertex, the two points of incidence and the intersection of the normals has two right angles. Therefore, $A + \theta = 180^\circ$.
2. The triangle formed by the internal ray and the two normals gives $r_1 + r_2 + \theta = 180^\circ$.
3. Subtract the first equation from the second to eliminate $\theta$. This gives $r_1 + r_2 - A = 0$.

$r_1 + r_2 = A$

All these angles are measured in degrees. The relation helps find the incidence at the second face before comparing it with the critical angle.

Worked example 4. A glass prism has a refracting angle of 60° and refractive index 1.524. Find the limiting angle of incidence at its first face for total internal reflection at the other face.

Answer: Let $i$ be the external incidence angle, $r$ the refracted angle at entry and $C$ the critical angle at the second face. At the limiting condition, the internal incidence at the second face equals $C$.

Formula: $C = \sin^{-1}(1/\mu)$, $r = A-C$ and $\sin i = \mu\sin r$, using the prism angle relation and Snell's law at entry from air.

Substitute: $C = \sin^{-1}(1/1.524) \approx 41.0083^\circ$ and $r = 60^\circ-41.0083^\circ \approx 18.9917^\circ$.

Then $i = \sin^{-1}[1.524\sin(18.9917^\circ)] \approx 29.73^\circ$. The limiting incidence is approximately 29.73°. At equality the emerging ray grazes the second face. For this path, a slightly smaller entry angle increases the second-face incidence above $C$, giving total internal reflection.

What happens in an equilateral prism?

An equilateral prism has three equal cross-section angles of 60°. Consider a crown-glass prism in air, with a ray entering normally through one sloping face and positioned so that it next meets the base.

The ray enters without bending. Geometry gives an incidence angle of 60° at the base. Since 60° exceeds the crown-glass critical angle of 41.14°, total internal reflection occurs there. The reflected ray then reaches the other sloping face normally and emerges without bending.

Draw a triangle with a horizontal base and a 60° apex above it. Send a ray normally through the left sloping face towards the base. Mark the base normal, the 60° incidence angle and the reflected ray emerging normally through the right sloping face.

Why can a 30°, 60°, 90° prism give different outcomes?

Orient this right-angled triangle with a horizontal base, a vertical left side and a hypotenuse sloping down towards the right. Put 60° at the upper-left vertex, 90° at the lower-left vertex and 30° at the right vertex.

A horizontal ray entering normally through the vertical side meets the hypotenuse at 60° to its normal. In crown glass surrounded by air, this exceeds 41.14°, so total internal reflection occurs. Follow the reflected ray to whichever face its entry position makes it reach next.

A vertical ray entering normally upwards through the base instead meets the same hypotenuse at 30° to its normal. Since 30° is below 41.14°, the ray is partly refracted into air. This encounter does not produce total internal reflection.

Draw the stated triangle twice. In one copy, show horizontal normal entry through the vertical side and 60° incidence at the hypotenuse. In the other, show vertical normal entry through the base and 30° incidence at the hypotenuse. Draw a normal at each encounter.

The decision method is therefore to find the internal incidence angle, identify the medium beyond that face, compare with the relevant critical angle, and only then draw reflection or refraction.

How does total internal reflection compare with a plane mirror?

A plane mirror is a mirror with a flat reflecting surface. Both a plane mirror and a totally reflecting prism can redirect light. In both cases, the reflected ray obeys the law of reflection, so its angle to the normal equals the incidence angle.

What is different about the reflection?

The advantage of total internal reflection is complete reflection at the internal boundary when its conditions are met. A prism can therefore provide an effective reflecting surface without a mirror coating on that face.

Keep the comparison local to the reflecting boundary. A prism also has entry and exit faces. Complete reflection at one internal face does not justify claiming that a whole optical instrument has no loss of light anywhere.

The limitation is the need to satisfy the medium and angle conditions. A prism cannot be treated as a mirror for every possible incoming ray. If a reflecting encounter occurs below the critical angle, part of the light can pass into the surrounding medium.

In a ray diagram, show the actual prism faces and normals. A 90° turn or a 180° reversal follows from a particular path through those faces, rather than from a general rule that every triangular prism produces the same deviation.

How do optical fibres use total internal reflection?

An optical fibre is a fine transparent guide that carries light along its length. Its central region is the core; the surrounding layer is the cladding. The core material has a higher refractive index than the cladding material.

How does light remain inside the core?

1. A light signal enters one end of the fibre at a suitable angle.
2. Inside the core, it reaches the core-cladding boundary from the optically denser side.
3. Its incidence angle at that boundary exceeds the critical angle, producing total internal reflection.
4. Repeated reflections guide the light along the fibre until it emerges at the other end.

The angle at entry and the incidence angle at the side boundary are different angles. The important comparison with C is made where the ray strikes the core-cladding interface. Saying merely that light has entered the fibre is not enough to establish the required path.

A curved fibre contains a zigzag ray path. The inner region is labelled with a higher refractive index and the surrounding region with a lower refractive index. Arrows show light entering, reflecting successively and leaving the other end.

See Fig. 9.14 in your NCERT textbook

What are the applications and practical limits?

Optical fibres are extensively used to transmit audio and video signals over long distances. Signals can be converted into light before transmission. Bundles of fibres also act as light pipes for visual examination of internal organs such as the oesophagus, stomach and intestines.

Decorative fibre lamps provide another application. Light from a lamp enters the fibres at one end and appears as bright dots at their free ends. Each fibre guides light along its own length.

Repeated total internal reflection causes no appreciable loss in the intensity of the light signal at the reflecting stages. Nevertheless, fibre materials must have very little absorption, meaning very little light energy is taken up by the material during transmission.

Even if a fibre is bent, light can travel along its length when the guiding conditions are maintained. Preserve the distinction between efficient guidance and a claim of absolutely lossless transmission. The quality of the material and the internal incidence angles both matter.

Which entry angles allow a light pipe to guide rays?

Worked example 5. A glass light pipe has a core refractive index of 1.68 and an outer covering of refractive index 1.44. Find the range of entry angles in air, measured from the pipe's axis, for total internal reflection inside the pipe.

Answer: Let $i$ be the angle in air, $r$ the refracted angle to the axis and $q$ the incidence at the side boundary. Let $n$ be the core index and $n_c$ the covering index. All angles below are in degrees.

Formula: $\sin C = n_c/n$, $q = 90^\circ-r$ and $\sin i = n\sin r$. At the limiting entry angle, $q=C$, giving $\sin i_{\mathrm{lim}} = n\cos C$.

Substitute: $C = \sin^{-1}(1.44/1.68) \approx 58.9973^\circ$, so the limiting angle within the core is $r = 90^\circ-C \approx 31.0027^\circ$.

Hence $i_{\mathrm{lim}} = \sin^{-1}[1.68\cos C] \approx 59.92^\circ$. Non-axial rays entering below approximately 59.92° undergo total internal reflection at the side boundary. The equality case is the critical limit; an exactly axial ray travels straight along the pipe.

Worked example 6. Repeat the light-pipe calculation for the same glass of refractive index 1.68 when its outer covering is removed and the glass is surrounded by air.

Answer: The core-to-air boundary now has $\sin C = 1/1.68$, so $C \approx 36.53^\circ$. At entry from air, Snell's law gives $\sin r = \sin i/1.68$.

Since $\sin i$ cannot exceed one, the refracted angle approaches at most 36.53°. Thus the incidence at the side boundary is at least approximately $q = 90^\circ-36.53^\circ = 53.47^\circ$, which exceeds the critical angle.

Every ray that enters the end face from air and reaches the side boundary therefore satisfies the total internal reflection condition. Entry angles may range from 0° up to, but not including, 90° to the axis. The axial ray travels straight; at 90° a ray is parallel to the entry face.

Glossary

• Interface — The boundary separating two media at which reflection or refraction can occur.
• Normal — A line perpendicular to a surface at the point where a ray strikes.
• Angle of incidence — The angle between the incident ray and the normal at the boundary.
• Angle of refraction — The angle between the refracted ray and the normal in the second medium.
• Optically denser medium — A medium with a higher refractive index than the other medium being compared.
• Optically rarer medium — A medium with a lower refractive index than the other medium being compared.
• Critical angle — The incidence angle in the denser medium for which refraction occurs along the boundary.
• Total internal reflection — Complete reflection back into the denser medium at a denser-to-rarer boundary when incidence exceeds the critical angle.
• Refractive index — A dimensionless ratio describing the relative speed of light in the media being compared.
• Angle of deviation — The change between a ray’s original incident direction and its final emergent direction.
• Hypotenuse face — The prism face corresponding to the side opposite the right angle in its triangular cross-section.
• Optical fibre — A fine transparent guide that carries light along its length through repeated total internal reflections.
• Core — The central light-guiding region of an optical fibre, with higher refractive index than its cladding.
• Cladding — The surrounding layer of an optical fibre, with lower refractive index than the core.

Common errors and misconceptions

• Misconception: Total internal reflection occurs whenever light travels from glass towards air. Correct: The incidence angle must also exceed the glass-air critical angle. The direction of travel supplies only one of the two necessary conditions.
• Misconception: At the critical angle, the incident ray makes 90° with the normal. Correct: The refracted ray makes 90° with the normal. The incident ray makes the critical angle with its normal inside the denser medium.
• Misconception: Total internal reflection starts when i equals C. Correct: At equality, the refracted ray grazes the interface. Total internal reflection requires i to be greater than C, so the equality case must be kept separate.
• Misconception: The incidence angle is measured from the prism face. Correct: It is measured from the normal to that face. Draw a fresh normal at every face encountered, especially when the ray changes direction inside a prism.
• Misconception: A right-angle prism must turn every incident ray through 90°. Correct: Its effect depends on entry direction and the internal path. The specified normal-entry arrangements produce either a 90° turn or a 180° reversal.
• Misconception: A higher refractive index means a larger critical angle. Correct: For comparison with the same rarer medium, μ = 1/sin C gives a smaller critical angle when the refractive index is higher.
• Misconception: Optical fibres transmit light with absolutely no loss anywhere. Correct: Total internal reflection gives no appreciable loss at the reflecting stages, but the material must also have very little absorption over the transmission distance.

Exam-style questions with model answers

Q1. State the two essential conditions for total internal reflection at a boundary between transparent media. [2 marks]

1. The incident light must travel from an optically denser medium towards an optically rarer medium.
2. Its angle of incidence at the boundary must be greater than the critical angle for that pair of media.

Q2. A ray approaches a water-air boundary from within water. Describe what happens below, at and above the critical angle C, where C is the water-air critical angle. [3 marks]

1. Below C, some light is reflected into water and some enters air. For oblique incidence, the refracted ray bends away from the normal.
2. At C, the refracted ray makes an angle of 90° with the normal and travels along the water-air interface.
3. Above C, the light undergoes total internal reflection back into water. No transmitted ray enters air in the ray description.

Q3. Let μ be the refractive index of a denser medium relative to a rarer medium, and C their critical angle. Starting from μ sin i = sin r, where i and r are the incidence and refraction angles, derive the relation between μ and C. Use sin 90° = 1. [3 marks]

1. At the critical angle, the angle of incidence i is C. The refracted ray travels along the boundary, making its angle of refraction r equal to 90°.
2. Substitute these limiting values into the given relation to obtain μ sin C = sin 90° = 1.
3. Divide by sin C to obtain μ = 1/sin C. The refractive index is a dimensionless ratio for the specified pair of media.

Q4. A ray enters normally through a shorter face of a 45°, 45°, 90° crown-glass prism in air and next strikes the hypotenuse face. The glass-air critical angle is 41.14°. Trace its path and give the final deviation. [5 marks]

1. The ray enters the shorter face along its normal. Its incidence angle at entry is 0°, so it passes into the glass without changing direction.
2. The prism geometry makes the internal incidence angle at the hypotenuse face 45°, measured from the normal to that face.
3. The ray approaches air from denser glass, and 45° exceeds the given critical angle of 41.14°. It therefore undergoes total internal reflection.
4. The reflected ray reaches the other shorter face normally and emerges into air without bending at this exit face.
5. The emergent direction is perpendicular to the incident direction. The angle of deviation is therefore 90°.

Q5. A ray enters normally through the hypotenuse face of a 45°, 45°, 90° crown-glass prism in air, away from the position leading to a vertex. It next encounters both shorter faces. The glass-air critical angle is 41.14°. Explain the path and final deviation. [5 marks]

1. The ray enters the hypotenuse face normally and continues straight into the glass. There is no change in its direction at entry.
2. It reaches the first shorter face at 45° incidence. Since the ray is in glass and 45° exceeds 41.14°, total internal reflection occurs.
3. The reflected ray meets the second shorter face at 45° incidence. The same denser-to-rarer direction and critical-angle comparison give a second total internal reflection.
4. After these two reflections, the ray returns normally to the hypotenuse face and passes out into air without bending there.
5. The outgoing ray travels opposite to the incoming direction, giving a deviation of 180°. It is parallel to, but need not retrace, the incident ray.

Q6. For an equilateral crown-glass prism in air, each cross-section angle is 60° and the glass-air critical angle is 41.14°. A ray enters normally through the left sloping face, strikes the horizontal base, then reaches the right sloping face. Explain these three encounters. [3 marks]

1. At entry through the left sloping face, the ray follows the normal. It therefore enters the crown glass without bending.
2. The equilateral geometry gives 60° incidence at the horizontal base. This exceeds 41.14°, and the ray approaches air from glass, so total internal reflection occurs.
3. The reflected ray meets the right sloping face normally. It leaves the prism without bending at that face, completing the specified three-face path.

Q7. Compare total internal reflection at a prism’s glass-air boundary with ordinary reflection from a plane mirror, stating the angle condition and the amount of light reflected. [2 marks]

1. Total internal reflection requires incidence from glass above the critical angle; ordinary mirror reflection has no such critical-angle requirement.
2. The qualifying internal boundary reflects all incident light, whereas ordinary mirror reflection does not return all incident light into the reflected ray.

Q8. An optical fibre has a core whose refractive index is higher than its cladding’s. A light signal enters at a suitable angle so that incidence at each core-cladding encounter exceeds the critical angle. Explain its guidance, state one application and explain why the material should have very little absorption. [4 marks]

1. At the core-cladding boundary, the light approaches from the optically denser core towards the optically rarer cladding, satisfying the direction condition.
2. The given incidence angles exceed the critical angle. Repeated total internal reflections therefore guide the light along the fibre to its other end.
3. Optical fibres can transmit audio and video signals over long distances by carrying the information in light signals.
4. Very little absorption is needed so the material takes up very little light energy during transmission. Efficient internal reflection alone does not eliminate absorption within the material.

Key takeaways

• Total internal reflection requires both denser-to-rarer travel and an incidence angle greater than the critical angle.
• At the critical angle, the refracted ray travels along the interface and makes 90° with the normal.
• The relation μ = 1/sin C connects relative refractive index with critical angle for the specified pair of media.
• Measure incidence and reflection angles from the normal at the actual face where the ray arrives.
• A 45°, 45°, 90° prism can give 90° deviation through one internal reflection or 180° deviation through two.
• For the standard right-angle prism paths in air, the glass-air critical angle must be less than 45°.
• Equilateral and 30°, 60°, 90° prism paths require explicit entry directions and separate checks at each boundary.
• Optical fibres guide light by repeated total internal reflection in a higher-index core surrounded by lower-index cladding.

Test yourself

Why does a large incidence angle not guarantee total internal reflection?

The ray must also approach the boundary from the optically denser medium towards the optically rarer medium.

Which angle equals 90° at the critical condition?

The angle of refraction equals 90°, so the refracted ray runs along the interface between the media.

What happens when incidence is exactly equal to the critical angle?

The refracted ray grazes the boundary. Total internal reflection requires incidence greater than the critical angle.

What is the unit of refractive index?

Refractive index has no unit because it is a ratio of quantities with the same units.

Where does light enter a 45°, 45°, 90° prism in the standard 180° reversing arrangement?

It enters normally through the hypotenuse face, then undergoes total internal reflection at both shorter faces.

Does a 180° deviation mean that a ray retraces its original line?

No. The emergent ray travels in the opposite direction but may be displaced from the incident ray’s line.

Why must the core have a higher refractive index than the cladding?

This makes light approach the core-cladding boundary from the optically denser side, allowing total internal reflection above the critical angle.

Why should optical-fibre material absorb very little light?

Absorption reduces the light energy travelling through the material, even when total internal reflection efficiently guides the ray.