Refraction of light through a glass block and a triangular prism | ICSE Class 10 Physics Notes

This note covers refraction, the change in light's direction on entering another transparent material obliquely; refractive index, a ratio describing relative light speeds; light passing through glass blocks and triangular prisms; apparent depth, the depth at which an object seems to lie; and multiple images produced by glass surfaces.

What happens when light meets another transparent medium?

A medium is a material through which light travels. A transparent medium allows light to pass through it. The boundary between two media is their interface. Air, water and glass provide familiar examples of media with different effects on travelling light.

A ray is a line representing the direction in which light travels. When a ray meets a boundary obliquely, it arrives at a slant rather than perpendicular to the surface. Its direction can change as it enters the second medium.

Definition: Refraction of light is the change in direction of an obliquely incident light ray when it passes from one transparent medium into another because its speed changes.

How do reflection and refraction occur together?

Reflection is the return of light into the medium from which it arrived. At an ordinary transparent boundary, part of the incident light is reflected and part enters the other medium. This simultaneous division is called partial reflection and refraction.

The incident ray approaches the boundary, the reflected ray returns into the first medium, and the refracted ray travels into the second. A transmitted ray is light that passes through a boundary rather than returning by reflection.

The normal is an imaginary line perpendicular to the surface at the point where a ray meets it. Draw this line before describing bending. “Towards the normal” and “away from the normal” refer to the ray's angle with this line, not its distance from the surface.

Why does the distinction matter?

Glass can transmit light and still produce a reflected image. Seeing through a glass plate therefore does not mean that all the incident light has passed through it. Conversely, a reflected image does not mean that the glass has stopped transmitting light.

Refraction explains why a pencil partly immersed in water appears displaced at the surface and why printed letters viewed through a thick glass slab appear raised. The objects remain in place; the light reaching the observer follows a changed path.

What are the laws of refraction and the rules for bending?

The angle of incidence, written i, is the angle between the incident ray and the normal. The angle of refraction, written r, is the angle between the refracted ray and the normal. Here angles are expressed in degrees, represented by °.

1. The incident ray, refracted ray and normal at the point of incidence all lie in the same plane.
2. For light of a given colour and a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction remains constant.

The second statement is Snell's law. The sine, written sin, is a trigonometric ratio; for an acute angle in a right-angled triangle, it equals the opposite side divided by the hypotenuse, the side opposite the right angle.

Let n₂₁ mean the refractive index of medium 2 with respect to medium 1, with light travelling from medium 1 into medium 2. For oblique incidence, with 0° < i < 90°:

n₂₁ = sin i / sin r

What do optically denser and optically rarer mean?

An optically denser medium has a larger refractive index and a lower speed of light than the medium with which it is compared. An optically rarer medium has a smaller refractive index and a higher light speed.

Normal incidence means arrival along the normal. The transmitted ray then continues undeviated, meaning without a change in direction, even when its speed changes. A boundary between media of equal refractive index also produces no bending of the transmitted ray.

Note: Optical density is not mass density, which is mass per unit volume. Kerosene is optically denser than water although its mass density is lower. Do not decide the bending direction from how heavy a material seems.

How does refractive index determine the speed of light?

The absolute refractive index, written n or µ (mu), compares light's speed in vacuum with its speed in a medium. A vacuum is a region without matter. Let c represent light's speed in vacuum and V its speed in the selected medium.

n = c / V

V = c / n

The SI unit of speed is metre per second, written m/s. SI means the International System of Units. Light travels fastest in vacuum, with c = 3 × 10⁸ m/s. In air its speed is only marginally less than in vacuum.

Refractive index has no unit: the two speeds in its ratio have the same units, which cancel. A larger index means a lower speed for the light being compared. Glass and water reduce the speed considerably compared with vacuum.

Which values should be used?

Use the index supplied for a particular material. “Glass” is not one unique optical material: a problem may supply 1.50 for its glass, while crown glass has the listed value 1.52. These inputs must not be silently exchanged during a calculation.

Worked example 1. Glass has refractive index 1.50. Calculate its light speed using c = 3 × 10⁸ m/s.

Formula: V = c/n. Substitute: V = (3 × 10⁸)/1.50. Answer: V = 2 × 10⁸ m/s, or 200,000,000 m/s. The speed is below the vacuum value.

Worked example 2. Water has refractive index 1.33. Calculate its light speed using c = 3 × 10⁸ m/s.

Formula: V = c/n. Substitute: V = (3 × 10⁸)/1.33. Answer: V ≈ 2.26 × 10⁸ m/s, or approximately 226,000,000 m/s. The symbol ≈ means approximately equal to; the calculated result has been rounded.

Worked example 3. Air has refractive index 1.0003. Calculate its light speed using c = 3 × 10⁸ m/s.

Formula: V = c/n. Substitute: V = (3 × 10⁸)/1.0003. Answer: V ≈ 2.9991 × 10⁸ m/s, or approximately 299,910,000 m/s. For simple calculations, air's speed is therefore commonly approximated by c.

Worked example 4. Diamond has refractive index 2.42. Calculate its light speed using c = 3 × 10⁸ m/s.

Formula: V = c/n. Substitute: V = (3 × 10⁸)/2.42. Answer: V ≈ 1.24 × 10⁸ m/s, or approximately 124,000,000 m/s. The index 2.42 describes a speed ratio, not a speed in metres per second.

How are relative refractive index, wavelength and frequency connected?

Let V₁ and V₂ be light's speeds in media 1 and 2, and n₁ and n₂ their absolute refractive indices. The small numerals identify the media. The index n₂₁ compares medium 2 with medium 1; reversing this order reverses the speed ratio.

n₂₁ = V₁ / V₂

Derivation: relative index from absolute indices

1. Write the speed in medium 1 as V₁ = c/n₁.
2. Write the speed in medium 2 as V₂ = c/n₂.
3. Substitute both expressions into n₂₁ = V₁/V₂ and cancel c.

n₂₁ = n₂ / n₁

Let n₁₂ denote the index of medium 1 with respect to medium 2. Reversing the two media gives the reciprocal relationship:

n₁₂ = 1 / n₂₁

Worked example 5. Water and crown glass have absolute indices 1.33 and 1.52. Find the index of crown glass relative to water and then its light speed. Use c = 3 × 10⁸ m/s. Treat water as medium 1 and crown glass as medium 2.

Formula: n₂₁ = n₂/n₁; V₁ = c/n₁; V₂ = V₁/n₂₁. Substitute: n₂₁ = 1.52/1.33; V₂ = [(3 × 10⁸)/1.33]/(1.52/1.33). Answer: n₂₁ ≈ 1.14 and V₂ ≈ 1.97 × 10⁸ m/s, or approximately 197,000,000 m/s. Keep unrounded ratios until the last step.

What changes in the light wave?

Wavelength, written λ and read “lambda”, is the distance between successive wave crests, or corresponding points in consecutive cycles. A crest is a wave's peak. The SI unit of wavelength is metre, written m.

Frequency, written f, is the number of oscillations, or complete repeating cycles, per second. The SI unit of frequency is hertz, written Hz. Light's speed, frequency and wavelength are related by:

V = fλ

f = V / λ

During refraction at a stationary boundary, frequency remains the same. Speed and wavelength decrease on entering an optically denser medium. They increase on entering an optically rarer medium. The unchanged frequency links the two changes.

Let λ₁ and λ₂ be wavelengths in media 1 and 2. Since the same f applies on both sides, dividing the two speed equations gives:

λ₂ / λ₁ = V₂ / V₁

Worked example 6. Monochromatic light, meaning light of a single frequency, has wavelength 589 nm in air and enters water of index 1.33. A nanometre, written nm, equals 10⁻⁹ m. Take air's speed as c = 3 × 10⁸ m/s. Find the frequency, speed and wavelength in water.

Formula: f = c/λ₁; V = c/n; λ₂ = V/f. Substitute: f = (3 × 10⁸)/(589 × 10⁻⁹); V = (3 × 10⁸)/1.33; λ₂ = (589 × 10⁻⁹)/1.33. Answer: f ≈ 5.09 × 10¹⁴ Hz, V ≈ 2.26 × 10⁸ m/s (approximately 226,000,000 m/s) and λ₂ ≈ 443 nm. Frequency is unchanged.

How can refraction through a glass block be traced experimentally?

A rectangular glass block, also called a rectangular glass slab, has opposite parallel faces. To trace a light path through it, use a white paper sheet, drawing board, drawing pins, a pencil, a ruler and four identical optical pins.

Optical pins are straight pins used to align a line of sight. The aim is to locate the ray entering the slab and the ray leaving it. The emergent ray is the ray that leaves the glass and travels into the surrounding air.

What is the tracing procedure?

1. Fix the paper to the board, place the slab centrally and trace its outline. Label the corners A, B, C and D so that AB and CD identify the opposite faces used.
2. Fix two pins vertically at positions E and F, with their joining line inclined to face AB.
3. Look through the opposite face. Fix pins G and H so that they and the images of pins E and F appear in one straight line.
4. Remove the pins and slab. Join E to F and extend the line to meet AB at O, the entry point.
5. Join G to H and extend the line to meet CD at O′, the exit point. Join O to O′ to obtain the ray inside the slab.
6. Draw normals at O and O′. Measure incidence, refraction and emergence angles with a protractor, an instrument for measuring angles.

The drawing labels the slab ABCD, incident ray EO, internal ray OO′ and emergent ray O′H. Normals are drawn at the two surfaces. A dotted continuation of the incident direction runs alongside the emergent ray.

See Fig. 9.10 in your NCERT textbook

How does this verify the laws?

Repeat the tracing for different incident directions, using the same glass and light colour. Record i, r, sin i, sin r and sin i/sin r. The measured ratios should be approximately constant, allowing for measurement error.

The incident ray, refracted ray and normal are represented in one plane on the paper. This illustrates the first law. The approximately constant sine ratio tests the second law; no fixed difference between i and r is predicted.

Why does a glass block produce lateral displacement?

At the air-to-glass surface, an oblique ray slows down and bends towards the normal. At the glass-to-air surface, it speeds up and bends away from the normal. These are separate refractions at two different boundaries.

For a rectangular block with air on both sides, the opposite faces are parallel. Their normals are therefore parallel too. The bending at the second face reverses the angular change produced at the first, so the emergent ray is parallel to the incident direction.

How is displacement different from deviation?

Lateral displacement is the perpendicular separation between the emergent ray and the original incident ray's forward continuation. The ray is shifted sideward slightly. Its final direction is restored, but its final line of travel is displaced.

Deviation means a change in direction. A ray can therefore have lateral displacement even when its net angular deviation, the difference between its initial and final directions, is zero. Do not confuse the distance between parallel lines with an angle between intersecting lines.

The angle of emergence, written e, is measured between the emergent ray and the normal at the exit face. For this parallel-sided slab with the same medium outside both faces:

e = i

What happens at normal incidence?

A ray entering perpendicular to the first face also reaches the opposite parallel face normally. It travels straight through, without lateral displacement. Its speed still decreases in glass and increases again on emerging into air.

Note: “Emerges undeviated” does not mean “undergoes no refraction at either face”. An oblique ray through a rectangular slab changes direction twice even though its final direction equals its initial direction.

How does a triangular glass prism change a ray's direction?

A triangular prism has two triangular bases and three rectangular lateral surfaces. The two faces through which the ray enters and leaves are its refracting faces. Unlike a slab's opposite faces, these prism faces are inclined to each other.

The angle of the prism, written A, is the angle between the two refracting faces. In the usual triangular cross-section, the opposite side is called the base of the prism diagram. Distinguish this side from the solid's two triangular end faces.

An oblique ray entering glass from air bends towards the first normal. On passing out from glass into air, it bends away from the second normal. The normals are inclined, so the two changes do not restore the incident direction.

The angle of deviation, written δ and read “delta”, is the angle between the forward continuation of the incident direction and the emergent ray. In the usual transmitted path through a glass prism in air, the deviation is towards the prism's base.

A triangular outline ABC contains ray EF. Incident ray PE reaches face AB at E, and emergent ray FS leaves face AC at F. Dotted normals and ray extensions show the incidence, refraction, emergence and deviation angles. The figure labels deviation D.

See Fig. 10.4 in your NCERT textbook

Derivation: the prism deviation relation

Let r₁ be the refraction angle inside glass at entry and r₂ the incidence angle inside glass at exit, each measured from its own normal. For the prism path:

A = r₁ + r₂

1. The change of direction at entry is i − r₁.
2. The change of direction at exit is e − r₂, so the total deviation is δ = (i − r₁) + (e − r₂).
3. Collect the terms and substitute r₁ + r₂ = A.

δ = i + e − A

The minus sign − denotes subtraction. The relation connects measurable angles; it does not make i and e equal for every path. Label each angle at its correct surface before using the relation.

How can prism refraction be investigated and compared with a block?

Prism tracing uses the same alignment principle as slab tracing. Place the prism on a sheet of paper with a triangular face against the paper, trace its outline and select an incident direction inclined to one refracting face.

What steps locate the transmitted path?

1. Place two pins along the selected incident line. These fix the direction of the ray approaching the first face.
2. View their images through the second refracting face. Align two more pins with the images so all appear in one straight line.
3. Remove the prism and pins. Extend the incident line to the first face and the emergent line back to the second face.
4. Join the entry and exit points to show the path through glass. Draw a separate normal at each point.
5. Mark i, r₁, r₂ and e. Extend the incident and emergent directions to locate δ, and measure A between the refracting faces.

Repeat for different incident directions. The measured angles can be checked against A = r₁ + r₂ and i + e = A + δ. Use the normal belonging to each face; a single normal cannot serve both inclined faces.

What is the key comparison?

The same bending rules operate in both objects. The difference comes from the arrangement of their surfaces. A prism is therefore not explained by a different law of refraction; apply Snell's law separately at each boundary.

A ray normal to a prism's first face enters without bending there. This fact alone does not establish its final direction: the second face has a different normal, so its behaviour must be considered separately.

Why do submerged objects appear raised and sticks appear bent?

Real depth is an object's actual perpendicular distance below a surface. Apparent depth is the depth at which it appears to lie when viewed through that surface. A submerged object seen from air can appear nearer the water surface than it really is.

Consider light travelling from the object in water towards an observer in air. At the water-air boundary, an oblique transmitted ray bends away from the normal. The observer receives rays whose directions differ from the paths they followed inside the water.

How does the ray construction locate the image?

1. Draw an object point below a horizontal water surface and two rays travelling from it towards the surface.
2. Draw normals at the points where the rays reach the boundary.
3. Show the transmitted rays bending away from these normals into air towards an observer.
4. Extend the outgoing rays backwards with dotted lines. Their backward extensions locate the apparent object point above the real point.

A virtual image is an apparent meeting point of backward ray extensions, rather than an actual meeting of the light rays. The raised underwater image is virtual; the object has not physically moved upwards.

The figure shows an observer above water, a real object point O below the surface and a higher apparent point O′. Solid rays emerge into air and dotted backward extensions locate the apparent point. Separate drawings show near-normal and oblique viewing.

See Fig. 9.10 in your NCERT textbook

A coin at the bottom of a bowl can become visible after water is added without moving the observer or coin. The coin appears slightly raised above its actual position because refraction changes the rays that can reach the eye.

Why does a partly immersed stick look bent?

Light from the part above water reaches the eye through air, while light from the submerged part crosses the water-air boundary. The underwater portion therefore appears displaced relative to the portion above water. The stick appears bent at the interface.

The same principle explains raised-looking letters beneath a thick glass slab. Apparent position depends on the media and viewing direction. Treat the explanation qualitatively: trace the light from the object towards the observer and distinguish its real position from its apparent position.

Why can a thick glass plate or mirror produce multiple images?

A thick glass plate has a front surface, facing the observer, and a rear surface farther away. Light can undergo partial reflection at both surfaces. Consequently, light reaching the observer can have followed more than one reflected path.

Which paths contribute?

Some light is reflected directly at the front air-glass boundary. Another part enters the glass by refraction, reaches the rear boundary and is reflected there. On returning to the front boundary, part of this light emerges into air.

Further parts can be reflected back inside the plate again before eventually emerging. These repeated reflections provide additional paths. Their backward extensions produce additional apparent images rather than a single image associated with only one reflecting surface.

Multiple images means more than one apparent image of the same object. The explanation requires both transmission and reflection: refraction allows light to enter and leave the plate, while reflections at different surfaces create the different returning paths.

What changes in a glass mirror?

An ordinary glass mirror has a reflecting coating behind the glass. A weak reflection can still occur at the front glass surface. Light transmitted through the glass is strongly reflected by the rear coating and returns towards the observer.

The rear coating produces the prominent image, while front-surface reflection and repeated internal reflections can produce fainter companion images. The glass thickness separates the surfaces involved, making their different optical paths significant. Do not describe these images as equally bright.

Draw parallel front and coated rear surfaces. Label air and glass. Show one ray reflected at the front, another refracted into glass and reflected from the coating, and a further path with an extra internal round trip before emergence.

Use arrows to distinguish entry, reflection and emergence. At each ordinary glass boundary, separate the transmitted branch from the reflected branch. A sketch that sends the whole ray along every branch would conceal the fact that the incident light is being divided.

This also connects the two observations made at a window: glass can allow a view through it and produce reflections at the same time. Transparency and partial reflection describe different portions of the arriving light.

Glossary

• Refraction — Change in direction of obliquely travelling light as it enters another transparent medium because its speed changes.
• Normal — Imaginary line perpendicular to a surface at the point where a light ray meets it.
• Angle of incidence — Angle between the incident ray and the normal at the point where the ray reaches a boundary.
• Angle of refraction — Angle between the refracted ray and the normal in the medium the ray enters.
• Absolute refractive index — Ratio of light's speed in vacuum to its speed in the selected transparent medium.
• Optically denser medium — Medium with a larger refractive index and lower light speed than another medium being compared.
• Wavelength — Distance between successive crests or equivalent points in consecutive cycles of a wave.
• Frequency — Number of complete oscillations in one second, unchanged when light undergoes refraction at a stationary boundary.
• Emergent ray — Ray that leaves a glass block or prism and enters the surrounding medium.
• Lateral displacement — Perpendicular separation between a slab's emergent ray and the forward continuation of its incident ray.
• Angle of deviation — Angle between the original forward incident direction and the direction of the emergent ray.
• Apparent depth — Depth at which a submerged object seems to lie when viewed through the surface.
• Virtual image — Apparent meeting point of backward ray extensions rather than a position where the light actually meets.
• Partial reflection — Reflection of part of the incident light while another part passes into the second medium.

Common errors and misconceptions

• Misconception: Measure refraction angles from the glass surface. Correct: Measure incidence, refraction and emergence angles from the appropriate normal, perpendicular to that surface.
• Misconception: A ray slows down only if it visibly bends. Correct: At normal incidence it can change speed without changing direction.
• Misconception: Optically denser means greater mass density. Correct: Optical density concerns refractive index and light speed; kerosene is optically denser than water despite lower mass density.
• Misconception: Frequency decreases as light enters glass. Correct: Frequency remains unchanged; reduced speed is accompanied by a reduced wavelength.
• Misconception: A parallel emergent ray proves no bending occurred. Correct: An oblique ray through a rectangular slab bends at both boundaries and emerges with lateral displacement.
• Misconception: A prism and slab restore the incident direction in the same way. Correct: Their face arrangements differ: the prism's inclined faces give a net angular deviation.
• Misconception: A submerged stick really bends and a coin rises when water is added. Correct: Refraction changes apparent positions while the objects remain in place.
• Misconception: All reflected images in a thick glass mirror have equal brightness. Correct: The rear coating produces the prominent image; other reflected paths can produce fainter companions.

Exam-style questions with model answers

Q1. State the two laws of refraction, including the conditions for the sine ratio to remain constant. [2 marks]

1. The incident ray, refracted ray and normal at the point of incidence lie in the same plane.
2. For a given colour and pair of media, sin i/sin r is constant for oblique incidence, where i and r are measured from the normal.

Q2. Light enters glass of refractive index 1.50. Given the vacuum speed 3 × 10⁸ m/s, calculate its speed in glass and explain the meaning of the index. [3 marks]

1. Refractive index is the ratio of vacuum speed c to medium speed V: n = c/V, hence V = c/n. Here n is the glass's absolute refractive index.
2. Substitution gives V = (3 × 10⁸)/1.50 = 2 × 10⁸ m/s, which is lower than the supplied vacuum speed.
3. The index means light travels 1.50 times as fast in vacuum as in this glass. It has no unit because it is a ratio of two speeds.

Q3. Explain the path of an oblique ray through opposite parallel faces of a rectangular glass slab with air on both sides. Distinguish lateral displacement from angular deviation. [5 marks]

1. At the first boundary, light travels from air into optically denser glass. It slows down and the transmitted ray bends towards the normal to that face.
2. The refracted ray then travels inside the glass to the opposite face. The normals at the two parallel slab faces are themselves parallel.
3. At the second boundary, the ray passes from glass into air. Its speed increases and the transmitted ray bends away from the exit normal.
4. The angular changes at the two boundaries cancel, so the emergent ray is parallel to the incident direction and the net angular deviation is zero.
5. The emergent ray is nevertheless shifted sideways. Lateral displacement is the perpendicular separation between it and the forward continuation of the original incident ray.

Q4. Describe how to trace a ray through a triangular glass prism using four pins and identify the angles needed to check δ = i + e − A. Here δ is deviation, i incidence, e emergence and A the prism angle. [5 marks]

1. Fix white paper on a board, place the prism on a triangular face and trace its outline. Set two pins on an incident line inclined to the first refracting face.
2. Look through the second refracting face and place two further pins so that they align with the images of the first two pins.
3. Remove the prism and pins. Extend each pair's joining line to its corresponding face and join the two intersection points to trace the internal ray.
4. Draw normals at the entry and exit points. Measure incidence i and emergence e from these normals, and measure A between the refracting faces.
5. Extend the original incident direction and the emergent ray to mark deviation δ. Compare measured δ with i + e − A, allowing for measurement error.

Q5. Light of wavelength 589 nm in air enters water of refractive index 1.33. Take air's speed as 3 × 10⁸ m/s and 1 nm = 10⁻⁹ m. Calculate its frequency, speed and wavelength in water, stating what remains unchanged. [4 marks]

1. Convert the given wavelength: 589 nm = 589 × 10⁻⁹ m. Frequency equals speed divided by wavelength, so f = (3 × 10⁸)/(589 × 10⁻⁹) ≈ 5.09 × 10¹⁴ Hz.
2. Frequency remains unchanged during refraction. Therefore the light in water has the same frequency as the incident light in air.
3. Using absolute index n = c/V, the speed in water is V = (3 × 10⁸)/1.33 ≈ 2.26 × 10⁸ m/s.
4. Since V = fλ, the water wavelength is λ = V/f ≈ 443 nm, using unrounded intermediate values. Both speed and wavelength decrease.

Q6. Explain why a coin under water appears raised when seen from air, and why a stick partly immersed in water appears bent. [3 marks]

1. Light from the submerged coin passes from optically denser water into air. Oblique transmitted rays bend away from the normal at the surface.
2. The backward extensions of rays reaching the eye locate a virtual image nearer the surface. The coin therefore appears slightly raised without actually moving.
3. Light from the submerged stick undergoes this refraction, while its part above water is seen through air. Their apparent positions do not line up, so the stick appears bent at the boundary.

Q7. Explain how a thick glass mirror with a rear reflecting coating can produce more than one image, and identify the prominent image. [3 marks]

1. A small portion of the arriving light reflects at the front air-glass surface, producing a weak front-surface image.
2. Another portion refracts into the glass, reflects strongly from the rear coating and emerges towards the observer. This path produces the prominent image.
3. Further internal reflections can occur before light leaves the glass. These additional reflected paths can produce fainter companion images, so all visible images need not have equal brightness.

Q8. What happens to the direction and speed of a transmitted ray entering glass normally from air? Explain using the normal. [2 marks]

1. Normal entry means the incident ray lies along the perpendicular to the surface, so the transmitted ray continues in that direction without bending.
2. Its speed decreases in the optically denser glass, even though the direction remains unchanged.

Key takeaways

• Measure incidence, refraction and emergence angles from the normal at the relevant surface, not from the surface itself.
• For a fixed colour and pair of media, Snell's law gives a constant ratio of the sines of incidence and refraction angles.
• Absolute refractive index compares vacuum speed with medium speed; a larger index corresponds to slower light.
• Frequency remains unchanged during refraction, while wavelength changes in the same ratio as the speed.
• A rectangular slab with air on both sides produces a parallel emergent ray and lateral displacement for oblique incidence.
• A triangular prism has inclined refracting faces, so its transmitted ray emerges with angular deviation.
• Submerged objects appear raised because refracted rays reaching the observer trace backwards to an apparent position nearer the surface.
• Partial reflections at different glass surfaces and repeated internal reflections can produce multiple images of one object.

Test yourself

From which line are angles of incidence and refraction measured?

They are measured from the normal, the line perpendicular to the boundary at the point of incidence.

What conditions make sin i/sin r constant?

The light colour and pair of media must remain fixed, with the angles measured from the normal for oblique incidence.

Does an undeviated ray necessarily retain its original speed?

No. A normally incident ray can change speed on entering glass while continuing in the same direction.

Which light-wave quantity remains unchanged during refraction?

Frequency remains unchanged; wavelength changes in proportion to the change in light speed.

Why are a slab's incident and emergent rays parallel in air?

The opposite faces are parallel, and their equal and opposite angular changes restore the original direction.

Why does the same cancellation not occur in a triangular prism's transmitted path?

The refracting faces and their normals are inclined, so the two refractions produce a net angular deviation.

Why is a raised underwater image called virtual?

The backward extensions of the emergent rays meet there; the actual light rays do not.

Which surface produces the prominent image in an ordinary rear-coated glass mirror?

The rear reflecting coating produces the prominent image after light has passed through the glass.