Refraction of Light Class 10 ICSE — Selina Chapter 4
Refraction of light class 10 ICSE physics begins with one observation: light changes direction when it crosses from one transparent medium into another. This chapter, Chapter 4 of the Selina Concise Physics book, explains that change. It covers Snell’s laws, refractive index, the prism, real and apparent depth, and total internal reflection. The numericals build directly on these ideas.
Download Refraction of Light At Plane Surfaces Class 10 ICSE Physics Chapter 4 PDF
Download the Refraction of Light At Plane Surfaces Class 10 ICSE Physics chapter PDF from the Selina Concise textbook to revise offline or print for practice. It is the full chapter as printed in the book — 33 pages, 16 sections, 78 questions.
What is inside Chapter 4
33 pages · 16 sections · 5 worked examples · 78 exercise questions
| Section | Title |
|---|---|
| 4.1 | REFRACTION OF LIGHT |
| 4.2 | LAWS OF REFRACTION |
| 4.4 | PRINCIPLE OF REVERSIBILITY OF THE PATH OF LIGHT |
| 4.5 | EXPERIMENTAL VERIFICATION OF LAWS OF REFRACTION AND DETERMINATION OF REFRACTIVE INDEX OF GLASS |
| 4.6 | REFRACTION OF LIGHT THROUGH A RECTANGULAR GLASS BLOCK |
| 4.7 | MULTIPLE IMAGES IN A THICK PLANE GLASS PLATE OR THICK MIRROR |
| 4.8 | PRISM |
| 4.9 | REFRACTION OF LIGHT THROUGH A GLASS PRISM |
| 4.10 | REAL AND APPARENT DEPTH |
| 4.12 | SOME CONSEQUENCES OF THE REFRACTION OF LIGHT |
| 4.13 | TRANSMISSION OF LIGHT FROM A DENSER MEDIUM (GLASS OR WATER) TO A RARER MEDIUM (AIR) AT DIFFERENT ANGLES OF INCIDENCE |
| 4.15 | RELATIONSHIP BETWEEN THE CRITICAL ANGLE AND THE REFRACTIVE INDEX (mu = 1/sin C) |
| 4.16 | TOTAL INTERNAL REFLECTION |
| 4.17 | TOTAL INTERNAL REFLECTION IN A PRISM |
| 4.18 | USE OF A TOTAL INTERNAL REFLECTING PRISM IN PLACE OF A PLANE MIRROR |
| 4.19 | SOME CONSEQUENCES OF TOTAL INTERNAL REFLECTION |
- EXERCISE-4(A): 27 questions
- MULTIPLE CHOICE TYPE: 7 questions
- MULTIPLE CHOICE TYPE: 4 questions
- EXERCISE-4(C): 8 questions
- MULTIPLE CHOICE TYPE: 5 questions
- EXERCISE-4(D): 24 questions
- MULTIPLE CHOICE TYPE: 3 questions
4.1 REFRACTION OF LIGHT
Refraction is the change in direction of light when it passes from one transparent medium to another. It is a surface phenomenon. At the boundary, light suffers partial reflection and partial refraction. A small part returns into the first medium, and the rest enters the second medium along a new path.
Three cases are observed. From a rarer to a denser medium, the ray bends towards the normal, so r \lt i and deviation \delta = i - r. From a denser to a rarer medium, it bends away from the normal, so r \gt i and \delta = r - i. At normal incidence, i = 0^{\circ}, r = 0^{\circ}, and \delta = 0^{\circ}.
One common mistake: optical density is not physical density. Kerosene floats on water, yet it is optically denser than water. Optical density depends on the speed of light in the medium, not on intermolecular separation.
Worked pattern: a ray enters glass from air at i = 40^{\circ} and refracts at r = 25^{\circ}. Since glass is denser, r \lt i. The deviation is \delta = 40^{\circ} - 25^{\circ} = 15^{\circ}.
4.2 LAWS OF REFRACTION
Snell’s laws govern refraction. First, the incident ray, the refracted ray, and the normal at the point of incidence all lie in the same plane. Second, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media. This constant is the refractive index:
\frac{\sin i}{\sin r} = {_1\mu_2} \tag{4.1}
The refractive index has no unit. It is defined by \mu = c/V, where c is the speed of light in vacuum or air (3 \times 10^8\ \text{m s}^{-1}) and V is the speed in the medium. Since V \lt c, the refractive index is always greater than 1. Other chapters in the Class 10 ICSE physics book list return to this relation.
Relative refractive index is {_1\mu_2} = V_1/V_2 = \mu_2/\mu_1. A ray passes undeviated in two edge cases: when i = 0^{\circ}, or when the two media have equal refractive indices. Frequency stays constant on refraction, but wavelength changes as \lambda' = \lambda/\mu.
Three factors affect refractive index: the nature of the medium, temperature, and colour or wavelength. Violet light has a higher refractive index than red light in the same medium.
Worked pattern: light travels at 2.4 \times 10^8\ \text{m s}^{-1} in a medium. Then \mu = (3 \times 10^8)/(2.4 \times 10^8) = 1.25. The medium slows light to 0.8 of its vacuum speed.
4.4 PRINCIPLE OF REVERSIBILITY OF THE PATH OF LIGHT
Light follows the same path in reverse. If a ray goes from medium 1 to medium 2 along AO to OB, then a ray sent along BO will retrace the path along OA. This is the principle of reversibility.
Two relations follow. The product {_1\mu_2} \times {_2\mu_1} = 1, so {_2\mu_1} = 1/{_1\mu_2}. Relative index can also be found from absolute indices. For example, _w\mu_g = \mu_g/\mu_w = (3/2)/(4/3) = 9/8.
Worked pattern: if _a\mu_m = 1.4, then _m\mu_a = 1/1.4 \approx 0.71. The reciprocal relation holds for every pair of media.
4.5 EXPERIMENTAL VERIFICATION OF LAWS OF REFRACTION AND DETERMINATION OF REFRACTIVE INDEX OF GLASS
The pin method verifies Snell’s laws in the lab. Place a rectangular glass block on a drawing board and trace its boundary. Draw a line AO at a chosen angle of incidence i and fix pins a and b on it. View from the opposite face and fix pins c and d so all four pins appear in a straight line. Join the dots to trace the refracted ray OB and the emergent ray BC.
Measure i and r, then compute \sin i / \sin r. This ratio is constant and equals the refractive index of glass. The circle method avoids measuring angles. Draw a circle centred at O. It cuts the incident ray at D and the refracted ray at E. Drop perpendiculars DF and EG onto the normal. Since OD = OE are radii of the same circle, \mu = DF/EG.
Worked pattern: in the circle method, measured lengths give DF = 3.4 cm and EG = 2.2 cm. Then \mu = 3.4/2.2 \approx 1.55. Repeat for i from 40° to 80° and average the results.
4.6 REFRACTION OF LIGHT THROUGH A RECTANGULAR GLASS BLOCK
A ray through a parallel-sided block bends twice. At the first surface, air to glass, it bends towards the normal. At the second surface, glass to air, it bends away from the normal. By reversibility, the angle of incidence equals the angle of emergence, so i = e. The emergent ray is parallel to the incident ray.
The emergent ray is shifted sideways. The perpendicular distance between the emergent ray and the original path is the lateral displacement x. It depends on the thickness of the block, the angle of incidence, and the refractive index. Violet light shifts more than red light because its refractive index is higher.
Worked pattern: for i = 30^{\circ} and glass of \mu = 1.5, Snell’s law gives \sin r = \sin 30^{\circ}/1.5 = 1/3, so r \approx 19.5^{\circ}. The emergent ray leaves parallel to the incident ray with a measurable lateral displacement.
4.7 MULTIPLE IMAGES IN A THICK PLANE GLASS PLATE OR THICK MIRROR
A thick plane mirror shows several images of one object. The first image is faint, formed by partial reflection at the front glass surface. Only about 4% of light reflects there. The second image is the brightest, formed by the strong reflection at the silvered back surface, which returns about 96% of the light.
Further images arise from repeated reflections inside the glass. Each one is fainter than the one before. The second image is brightest because it comes from the strongest single reflection.
4.8 PRISM
A prism is a transparent medium bounded by five plane surfaces with a triangular cross section. Two faces are identical parallel triangles, and three faces are rectangular. The triangle ABC is the principal section, and its angle A is the angle of prism.
The two rectangular faces that meet at the refracting edge are the refracting surfaces. The third rectangular face is the base. In ray diagrams, a prism is shown only by its principal section.
4.9 REFRACTION OF LIGHT THROUGH A GLASS PRISM
A monochromatic ray through a prism refracts at two inclined faces. At the first face it bends towards the normal; at the second it bends away. Both bends direct the ray towards the base of the prism.
Two key results connect the angles. The refracted angles inside the prism add to the prism angle: r_1 + r_2 = A. The deviation is \delta = (i_1 + i_2) - A.
| Property | Rectangular glass slab | Prism |
|---|---|---|
| Faces | Two parallel faces | Two inclined faces |
| Emergent ray | Parallel to incident ray | Deviated towards the base |
| Shift | Lateral displacement | No lateral displacement |
| Key relation | i = e | \delta = (i_1 + i_2) - A |
Deviation depends on four factors: the angle of incidence, the material or refractive index of the prism, the angle of prism, and the colour of light. Violet deviates most, red least.
The i–\delta curve shows deviation first falling, then rising. At minimum deviation, i_1 = i_2, the refracted ray is parallel to the base, and \delta_{min} = 2i - A.
Worked pattern: an equilateral prism has A = 60^{\circ}. At minimum deviation the angle of incidence is 50^{\circ}. Then \delta_{min} = 2 \times 50^{\circ} - 60^{\circ} = 40^{\circ}.
4.10 REAL AND APPARENT DEPTH
An object in a denser medium looks raised when viewed from a rarer medium. The emergent ray bends away from the normal, so the eye sees the object at a smaller depth. This idea appears often in the Class 10 ICSE syllabus.
The relations are _a\mu_m = \text{real depth}/\text{apparent depth}, apparent depth = real depth / \mu, and shift = real depth × (1 - 1/\mu). These hold only when viewing from vertically above.
The shift depends on the refractive index, the thickness of the medium, and the wavelength of light. Violet light gives more shift than red light. The same effect explains why a stick in water looks bent.
Worked pattern: a coin at 9 mm depth in water with \mu = 4/3 appears raised by 9 \times (1 - 3/4) = 2.25 mm. Its apparent depth is 9/(4/3) = 6.75 mm.
4.12 SOME CONSEQUENCES OF THE REFRACTION OF LIGHT
Many daily phenomena come from refraction. Stars twinkle because air density changes with temperature. The sun is visible a few minutes before sunrise and after sunset. A coin hidden behind the edge of a vessel becomes visible when water is poured in. Print under a glass slab looks raised, and a tank appears shallower than it is.
Worked pattern: a pond with real depth 4.0 m and \mu = 4/3 has apparent depth 4.0/(4/3) = 3.0 m. The pond looks 1.0 m shallower than it really is.
4.13 TRANSMISSION OF LIGHT FROM A DENSER MEDIUM (GLASS OR WATER) TO A RARER MEDIUM (AIR) AT DIFFERENT ANGLES OF INCIDENCE
When light travels from a denser to a rarer medium, the refracted ray bends away from the normal. The angle of refraction exceeds the angle of incidence. Three cases arise as the angle of incidence grows.
For small i, a weak reflected ray and a strong refracted ray appear, with r \gt i. As i grows, the refracted ray weakens. At i = C, the critical angle, r reaches 90^{\circ}. For i \gt C, no refracted ray exists, and the light is totally reflected.
The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90^{\circ}. For glass, C \approx 42^{\circ}; a ray at 30^{\circ} refracts with r \gt 30^{\circ}, while a ray at 42^{\circ} grazes the surface.
4.15 RELATIONSHIP BETWEEN THE CRITICAL ANGLE AND THE REFRACTIVE INDEX (mu = 1/sin C)
From Snell’s law at the critical angle, _g\mu_a = \sin C / \sin 90^{\circ} = \sin C. Since _a\mu_g = 1/_g\mu_a, we get _a\mu_g = 1/\sin C = \operatorname{cosec} C. In this ICSE chapter, this is the standard route from refractive index to critical angle.
Common critical angles include water ≈ 49°, glass ≈ 42°, flint glass ≈ 39°, and diamond ≈ 24°.
The critical angle depends on two factors. It is least for violet light and increases with wavelength. It also increases as temperature rises, because refractive index falls with temperature.
Worked pattern: a medium with \mu = 1.6 has \sin C = 1/1.6 \approx 0.625, so C \approx 39^{\circ}. A smaller refractive index gives a larger critical angle.
4.16 TOTAL INTERNAL REFLECTION
Total internal reflection is the complete return of light into the denser medium. It occurs when light in a denser medium strikes the boundary with a rarer medium at an angle of incidence greater than the critical angle.
Two conditions are essential. The light must travel from a denser to a rarer medium. The angle of incidence must exceed the critical angle. Then 100% of the energy is reflected, with no loss.
This perfect reflection is why a total reflecting prism replaces a plane mirror in periscopes, binoculars, and some cameras. A laser pen can demonstrate the effect, but never look directly into the beam. You can check the physics syllabus on the official CISCE website.
4.17 TOTAL INTERNAL REFLECTION IN A PRISM
A 45-90-45 prism, or total reflecting prism, has three actions. It can deviate a ray through 90°, used in periscopes. It can deviate a ray through 180°, used in binoculars and cameras. It can erect an inverted image without deviation, used in slide projectors.
An equilateral prism, with each angle 60°, deviates a ray through 60° by total internal reflection. A 30-90-60 prism deviates through less than 60°, except when light enters the hypotenuse below the foot of the perpendicular. Then the deviation exceeds 60°.
4.18 USE OF A TOTAL INTERNAL REFLECTING PRISM IN PLACE OF A PLANE MIRROR
A total reflecting prism gives brighter images than a plane mirror. Total internal reflection returns 100% of the light. A plane mirror absorbs and refracts part of the light, and its silvering deteriorates with age. Students working through the full list of Class 10 ICSE books will meet this comparison again in optics.
| Total internal reflection | Reflection from a plane mirror |
|---|---|
| Needs a denser-to-rarer boundary with i \gt C | Works from any medium at any angle |
| Entire light is reflected; no energy loss | Part is refracted and absorbed |
| Image stays bright with long use | Image fades as silvering ages |
This is why periscopes and binoculars use the prism instead of a mirror.
4.19 SOME CONSEQUENCES OF TOTAL INTERNAL REFLECTION
Total internal reflection explains several daily sights. A mirage appears as a pool of water on a hot road or in a desert. An empty test tube in water shines like a mirror at certain angles. Cracks in glass look bright. A diamond sparkles because of repeated internal reflections.
Optical fibres transmit light over long distances with only absorption losses. Each internal reflection keeps the light inside the fibre. In the Class 10 ICSE syllabus, these applications tie the theory to real devices.
What to practise from Refraction of Light Class 10 ICSE
In this Class 10 ICSE physics chapter, the numericals map to four question types: speed and refractive index problems using \mu = c/V, real and apparent depth problems, prism deviation problems using \delta = (i_1 + i_2) - A, and critical angle problems using \mu = 1/\sin C.
- EXERCISE-4(A): 27 questions on refraction, Snell’s laws, and the glass block
- Multiple choice type (after Exercise 4(A)): 7 questions
- Multiple choice type (after the prism): 4 questions
- EXERCISE-4(C): 8 questions on real and apparent depth
- Multiple choice type (after Exercise 4(C)): 5 questions
- EXERCISE-4(D): 24 questions on critical angle and total internal reflection
- Multiple choice type (after Exercise 4(D)): 3 questions
Refraction of Light Class 10 ICSE frequently asked questions
What is refraction of light in class 10 ICSE physics?
Refraction is the change in direction of light when it passes from one transparent medium to another. It is a surface phenomenon caused by the change in speed of light. At the boundary, part of the light is reflected and part is refracted.
Why does a ray of light bend towards the normal when it enters glass from air?
Glass is optically denser than air, so light slows down as it enters. When light slows down, it bends towards the normal. The angle of refraction becomes smaller than the angle of incidence.
How is the critical angle related to the refractive index of a medium?
The critical angle C satisfies \mu = 1/\sin C. A higher refractive index gives a smaller critical angle. Diamond, with \mu = 2.41, has a critical angle of about 24°.
What is the angle of minimum deviation for an equilateral glass prism?
At minimum deviation, \delta_{min} = 2i - A. For an equilateral prism, A = 60^{\circ}. If the angle of incidence is 50°, the minimum deviation is 40°.
Why is a total reflecting prism preferred over a plane mirror in a periscope?
Total internal reflection returns 100% of the light, while a plane mirror loses some light to absorption and refraction. The prism gives a brighter image that does not fade with use, unlike a mirror whose silvering deteriorates.
Reference: ICSE Class 10 Physics textbook material, chapter 4: Refraction of Light At Plane Surfaces — CISCE-aligned.